number theory

Total Gadha’s Complete Book of NUMBER SYSTEM http://www.totalgadha.com TYPES OF NUMBERS Natural Numbers The group of numbers starting from 1 and including 1, 2, 3, 4, 5, and so on, are known as natural numbers. Zero, negative numbers, and decimals are not included this group. If n is an odd natural number, what is the highest number that always divides n × (n2 – 1)? Answer: n × (n2 – 1) = (n – 1) × n × (n + 1), which is a product of three consecutive numbers. Since n is odd, the numbers (n – 1) and (n + 1) are both even. As they are two consecutive even numbers one of these numbers will be a multiple of 2 and the other will be a multiple of 4. Hence, their product is a multiple of 8. Since one out of every three consecutive numbers is a multiple of 3, one of the three numbers will be a multiple of three. Hence, the product of three numbers will be a multiple of 8 × 3 = 24. Hence, the highest number that always divides n × (n2 – 1) is 24. The product of n consecutive natural numbers is always divisible by n!, where n! = 1 × 2 × 3 × 4 × 5…. × n For every natural number n, the highest number that n × (n2 – 1) × (5n + 2) is always divisible by is (a) 6 (b) 24 (c) 36 (d) 48 Answer: Case 1: If n is odd, n × (n2 – 1) is divisible by 24 as proved in the earlier question. Case 2: If n is even, both (n – 1) and (n + 1) are odd. Since product of three consecutive natural numbers is always a multiple of 3 and n is even, the product n × (n2 – 1) is divisible by 6. Since n is even 5n is even. If n is a multiple of 2, 5n is a multiple of 2 and hence 5n + 2 is a multiple of 4. If n is a multiple of 4, 5n + 2 is a multiple of 2. Hence, the product n × (5n + 2) is a multiple of 8. Hence, the product n × (n2 – 1) × (5n + 2) is a multiple of 24. Prove that (2n)! is divisible by (n!)2. Answer: (2n)! = 1 × 2 × 3 × 4 × … × (n – 1) × n × (n + 1) × … × 2n = (n)! × (n + 1) × (n + 2) × … × 2n. Since (n + 1) × (n + 2) × … × 2n is a product of n consecutive numbers, it is divisible by n!. Hence, the product (n)! × (n + 1) × (n + 2) × … × 2n is divisible by n! × n! = (n!)2. Sum of first n natural numbers = n(n + 1) 2 n(n + 1)(2n + 1) 6 2 ⎛ n(n + 1) ⎞ ⎜ ⎟ 2 ⎝ ⎠ Sum of squares of first n natural numbers = Sum of cubes of first n natural numbers = Number of single-digit natural numbers: 1- 9 = 9 Number of two-digit natural numbers: 10- 99 = 90 Number of three-digit natural numbers: 100- 999 = 900 Square of every natural number can be written in the form 3n or 3n + 1. Square of every natural number can be written in the form 4n or 4n + 1. http://www.totalgadha.com If you write 1st 252 natural numbers in a straight line, how many times do you write the digit 4? Answer: In the 1st 99 natural numbers, digit 4 comes 20 times. Similarly, from 100 to 199, digit 4 comes 20 times. Now from 200 to 299, digit 4 comes again 20 times out of which we need to subtract 5 numbers (254, 264, 274, 284 and 294). Therefore, total number of times that we write the digit 4 = 20 + 20 + 20 − 5 = 55. If a book has 252 ages, how many digits have been used to number the pages? Answer: From page number 1 to age number 9, we will use 1 digit per page ⇒ digits used = 9. From page number 10 to age number 99, we will use 2 digits per page ⇒ digits used = 2 × 90 = 180. From page number 100 to age number 252, we will use 3 digits per page ⇒ digits used = 3 × 153 = 459. Therefore, total number of digits used = 9 + 180 + 459 = 648 There are three consecutive natural numbers such that the square of the second minus twelve times the first is three less than twice the third. What is the largest of the three numbers? Answer: Let the consecutive natural number be n, n + 1, n + 2. ⇒ (n + 1)2 − 12n = 2(n + 2) − 3. Solving, we get n = 12 and n + 2 = 14. What is the smallest natural number which is cube of a natural number and fourth power of a different natural number? Answer: Let N = x3 and N = y4. Therefore, N will contain 12th power (LCM of 3 and 4) of a natural number. Therefore, N = a12 = (a4)3 = (a3)4. The smallest such number is 212 = 4096. 1 and 8 are the first two natural numbers for which 1 + 2 + 3 + ... + n is a perfect square. Which number is the 4th such number? Answer: 1 + 2 + 3 + … + n = n(n + 1) 2 = M2 (say) ⇒ n(n + 1) = 2 M2 Now n and n + 1 will have no factor in common. Since RHS is twice the square of a natural number, one of n and n + 1 will be twice of a perfect square and the other will be a perfect square. As twice of a perfect square will be even, the other square will be odd. We start investigating the odd squares and their neighbours. The fourth such numbers we get is 288 × 289. Let N = 999 999 999 999 999 999. How many 9’s are there in N2? Answer: N2 = (1018 − 1)2 = 1036 + 1 − 2 × 1018 = 100000...00001 − 2000..0000 = 99999..9998000000...0001    35 zeroes 18 zeroes 17 9' s Whole Numbers All Natural Numbers plus the number 0 are called as Whole Numbers. Integers All Whole Numbers and their negatives are included in this group. For how many integers n is n4 + 6n < 6n3 + n2? http://www.totalgadha.com Answer: n4 + 6n − 6n3 − n2 < 0 ⇒ n (n2 − 1)(n − 6) < 0. n cannot be equal to 1 or 0 because LHS becomes 0. Now n2 − 1 will always be positive, therefore, n (n − 6) should be negative ⇒ n = 2, 3, 4 and 5. Find the sum of all two-digit positive integers which exceed the product of their digits by 12. Answer: Let the two-digit integer be ab. Therefore, 10a + b = ab + 12 ⇒ (a − 1)(10 − b) = 2 ⇒ numbers are 28 or 39. For which integer n is 28 + 211 + 2n is a perfect square? Answer: In order to write the above expression in the form (a + b)2 = a2 + 2ab + b2, we note that 28 = (24)2 and 211 = 2 × 24 × 26. Therefore, we need the square of 26 ⇒ 2n = (26)2 = 212 ⇒ n = 12. Find the smallest positive integer n for which (22 − 1)(32 − 1)(42 − 1)… (n2 − 1) is a perfect square. Answer: Nth term = (n2 − 1) = (n + 1)(n − 1) ⇒ series = 1 × 3 × 2 × 4 × 3 × 5 … × (n − 2) × (n) × (n − 1) × (n + 1) = 2 n (n + 1) × k2 because all the other terms are squared. The fist value of n which 2 n (n + 1) as perfect square is n = 8. Twenty cubical blocks are arranged as shown. First, 10 are arranged in a triangular pattern; then a layer of 6, arranged in a triangular pattern, is centered on the 10; then a layer of 3, arranged in a triangular pattern, is centered on the 6; and finally one block is centered on top of the third layer. The blocks in the bottom layer are numbered 1 through 10 in some order. Each block in layers 2, 3 and 4 is assigned the number which is the sum of the numbers assigned to the three blocks on which it rests. Find the smallest possible number which could be assigned to the top block. Answer: Let the numbers on the bottom 10 cubes be a, b, c, d, e, f, g, h, I, and j, as shown in the figure below: The sum on the top cube will come out to be a + g + j + 3(b + c + d + f + h + i) + 6e. Giving the lowest value to highest occurring number (e = 1) and highest values to lowest occurring numbers (a = 8, g = 9, j = 10), we get the minimum value = 114. Rational Numbers http://www.totalgadha.com Any number that can be expressed as a ratio of two integers is called a rational number. This group contains decimal that either do not exist (as in 6 which is 6/1), or terminate (as in 3.4 which is 34/10), or repeat with a pattern (as in 2.333... which is 7/3). Express 0.212121..in rational form. Answer: Let A = 0.21212121 ⇒ 100A = 21.212121.. = 21 + A ⇒ A = 21 99 . Rule: To express a recurring fraction in rational form, write the recurring digits once in the numerator and write as many 9s in the denominator as are the number of recurring digits. For example, 0.abcabcabc.. = abc 999 and 0.abcdabcdabcd.. = abcd 9999 Express 2.1434343…in rational form. Answer: Let R = 2.1434343.. ⇒ 10R = 21.43434343…= 21 + 43 99 ⇒R= 2122 990 . Rule: To write a fraction, which has both recurring and non-recurring parts, in a rational form, do the following steps: Numerator: (Number formed by writing all the digits once) − (Number formed by writing all the nonrecurring part once) = 2143 − 21 = 2122. Denominator: Number of 9s equal to number of recurring digits followed by number of zeroes equal to non-recurring digits after the decimal. What is the value of Answer: 22004 + 22001 22003 − 22000 ? 22004 + 22001 22001(23 + 1) 18 = = 7 22003 − 22000 22000(23 − 1) Irrational Numbers Any number that can not be expressed as the ratio of two integers is called an irrational number (imaginary or complex numbers are not included in irrational numbers). These numbers have decimals that never terminate and never repeat with a pattern. Examples include pi, e, and √2. 2 + √3, 5 - √2 etc. are also irrational quantities called Surds. 1 as a fraction whose denominator is rational. 5 + 6 − 11 1 5 + 6 + 11 = 5 + 6 − 11 ( 5 + 6 − 11) × ( 5 + 6 + 11) Express the value of Answer: = 5 + 6 + 11 = [( 5 + 6) 2 − ( 11) 2 ] If p = 8+ 7 and q = 8− 7 5 + 6 + 11 30( 5 + 6 + 11) = 60 2 30 8− 7 , then the value of p2 + pq + q2 is 8+ 7 http://www.totalgadha.com 8+ 7 ( 8 + 7 )2 = = ( 8 + 7)2 = 15 + 2 56 8− 7 ( 8 − 7)( 8 + 7) Answer: p = Similarly, q = ( 8 − 7)2 = 15 − 2 56 p2 + pq + q2 = (15 + 2 56)2 + (15)2 − (2 56)2 + (15 − 2 56)2 = 675 + 224 = 899 Real Numbers This group is made up of all the Rational and Irrational Numbers. The ordinary number line encountered when studying algebra holds real numbers. Imaginary Numbers These numbers are formed by the imaginary number i (i = −1 ). Any real number times i is an imaginary number. Examples include i, 3i, −9.3i, and (pi)i. Now i2 = −1, i3 = i2 × i = −i, i4 = 1. What is the value of i 4 + i 6 + i 8 + i10 + i12 ? i14 + i16 + i18 + i 20 + i 22 Answer: i4 = 1, i6 = i4 × i2 = −1, i8 = 1, i10 = −1, and so on. i 4 + i 6 + i 8 + i10 + i12 1−1+1−1+1 = = −1 Hence, 14 16 18 20 22 i +i +i +i +i −1 + 1 − 1 + 1 − 1 Complex Numbers A Complex Numbers is a combination of a real number and an imaginary number in the form a + bi. a is called the real part and b is called the imaginary part. Examples include 3 + 6i, 8 + (−5)i, (often written as 8 - 5i). Note: a number in the form 1 is written in the form of a complex number by multiplying both a + ib numerator and denominator by the conjugate of a + ib, i.e. a – ib. Hence, a − ib 1 a − ib a ib = = 2 = 2 − 2 , which is in the form p + iq. 2 2 (a + ib)(a − ib) a + b a + ib a +b a + b2 1+ i 7 ) is 1− i The value of ( Answer: (1 + i) 2 1 + i 2 + 2i 1+ i = = = 2i/2 = i 1 − i2 1 − i (1 − i )(1 + i) http://www.totalgadha.com 1+ i 7 ) = (i)7 = −i 1− i Hence, ( Prime Numbers All the numbers that have only two divisors, 1 and the number itself, are called prime numbers. Hence, a prime number can only be written as the product of 1 and itself. The numbers 2, 3, 5, 7, 11…37, etc. are prime numbers. Note: 1 is not a prime number. EXAMPLE If x2 – y2 = 101, find the value of x2 + y2, given that x and y are natural numbers. Answer: x2 – y2 = (x + y)(x – y) = 101. But 101 is a prime number and cannot be written as product of two numbers unless one of the numbers is 1 and the other is 101 itself. Hence, x + y = 101 and x – y = 1. Æ x = 51, y = 50. Æ x2 + y2 = 512 + 502 = 5101. What numbers have exactly three divisors? Answer: The squares of prime numbers have exactly three divisors, i.e. 1, the prime number, and the square itself. For how many prime numbers p, is p2 + 3p − 1 a prime number? Answer: When p = 3, the expression gives a prime number (17). When p is not equal to 3, p2 will be of the form 3k + 1 as every square number is of the form 3n or 3n + 1. Therefore, p2 + 3p − 1 = 3k + 1 + 3p − 1 ⇒ a multiple of 3. Therefore, for only p = 3, do we get a prime number from the expression. The number of positive integers n in the range 12 ≤ n ≤ 40 such that the product (n − 1) (n − 2) (n − 3)..3⋅2⋅1 is not divisible by n is (CAT 2003) Answer: The product (n − 1) (n − 2) (n − 3)..3⋅2⋅1 will not be divisible by n only when this product does not contain factors of n, i.e. n is a prime number. The prime numbers in the range given are 13, 17, 19, 23, 29, 31, and 37. 7 numbers in all. To find whether a number N is prime or not Find the root R (approximate) of the number N, i.e. R = N . Divide N by every prime number less than or equal to R. If N is divisible by at least one of those prime numbers it is not a prime number. If N is not divisible by any of those prime numbers, it is a prime number. Odd and Even Numbers All the numbers divisible by 2 are called even numbers whereas all the numbers not divisible by 2 are called odd numbers. 2, 4, 6, 8… etc. are even numbers and 1, 3, 5, 7.. etc. are odd numbers. http://www.totalgadha.com (Even)Odd = Even Even × Odd = Even Even × Even = Even Odd × Odd = Odd (Odd)Even × (Even)Odd = Even (Odd)Even + (Even)Odd = Odd Remember! Odd + Odd = Even Even + Even = Even Odd + Even = Odd (Odd)Even = Odd REMAINDERS Suppose the numbers N1, N2, N3… give quotients Q1, Q2, Q3… and remainders R1, R2, R3..., respectively, when divided by a common divisor D. Therefore N1 = D × Q1 + R1, N2 = D × Q2 + R2, N3 = D × Q3 + R3.. and so on. Let P be the product of N1, N2, N3… Therefore, P = N1N2N3.. = (D × Q1 + R1)(D × Q2 + R2)(D × Q3 + R3).. = D × K + R1R2R3... where K is some number ---- (1) Î In the above equation, since only the product R1R2R3… is free of D, therefore the remainder when P is divided by D is the remainder when the product R1R2R3… is divided by D. Let S be the sum of N1, N2, N3… Therefore, S = (N1) + (N2) + (N3) +... = (D × Q1 + R1) + (D × Q2 + R2) + (D × Q3 + R3).. = D × K + R1 + R2 + R3… where K is some number--- (2) Î Hence the remainder when S is divided by D is the remainder when R1 + R2 + R3 is divided by D. What is the remainder when the product 1998 × 1999 × 2000 is divided by 7? Answer: the remainders when 1998, 1999, and 2000 are divided by 7 are 3, 4, and 5 respectively. Hence the final remainder is the remainder when the product 3 × 4 × 5 = 60 is divided by 7. Therefore, remainder = 4 What is the remainder when 22004 is divided by 7? Answer: 22004 is again a product (2 × 2 × 2... (2004 times)). Since 2 is a number less than 7 we try to convert the product into product of numbers higher than 7. Notice that 8 = 2 × 2 × 2. Therefore we convert the product in the following manner- 22004 = 8668 = 8 × 8 × 8... (668 times). The remainder when 8 is divided by 7 is 1. Hence the remainder when 8668 is divided by 7 is the remainder obtained when the product 1 × 1 × 1... is divided by 7. Therefore, remainder = 1 What is the remainder when 22006 is divided by 7? Answer: This problem is like the previous one, except that 2006 is not an exact multiple of 3 so we cannot convert it completely into the form 8x. We will write it in following manner- 22006 = 8668 × 4. Now, 8668 gives the remainder 1 when divided by 7 as we have seen in the previous problem. And 4 gives a remainder of 4 only when divided by 7. Hence the remainder when 22006 is divided by 7 is the remainder when the product 1 × 4 is divided by 7. Therefore, remainder = 4 What is the remainder when 2525 is divided by 9? http://www.totalgadha.com Answer: Again 2525 = (18 + 7)25 = (18 + 7)(18 + 7)...25 times = 18K + 725 Hence remainder when 2525 is divided by 9 is the remainder when 725 is divided by 9. Now 725 = 73 × 73 × 73.. (8 times) × 7 = 343 × 343 × 343... (8 times) × 7. The remainder when 343 is divided by 9 is 1 and the remainder when 7 is divided by 9 is 7. Hence the remainder when 725 is divided by 9 is the remainder we obtain when the product 1 × 1 × 1... (8 times) × 7 is divided by 9. The remainder is 7 in this case. Hence the remainder when 2525 is divided by 9 is 7. What is the remainder when 32 3232 is divided by 7? SOME SPECIAL CASES: When both the dividend and the divisor have a factor in common. Let N be a number and Q and R be the quotient and the remainder when N is divided by the divisor D. Hence, N = Q × D + R. Let N = k × A and D = k × B where k is the HCF of N and D and k > 1. Hence kA = Q × kB + R. Let Q1 and R1 be the quotient and the remainder when A is divided by B. Hence A = B × Q1 + R1. Putting the value of A in the previous equation and comparing we getk(B × Q1 + R1) = Q × kB + R Î R = kR1. Hence to find • • • the remainder when both the dividend and the divisor have a factor in common, Take out the common factor (i.e. divide the numbers by the common factor) Divide the resulting dividend (A) by resulting divisor (B) and find the remainder (R1). The real remainder R is this remainder R1 multiplied by the common factor (k). What the remainder when 296 is divided by 96? The common factor between 296 and 96 is 32 = 25. Removing 32 from the dividend and the divisor we get the numbers 291 and 3 respectively. The remainder when 291 is divided by 3 is 2. Hence the real remainder will be 2 multiplied by common factor 32. Remainder = 64 The concept of negative remainder http://www.totalgadha.com 15 = 16 × 0 + 15 or 15 = 16 × 1 – 1. The remainder when 15 is divided by 16 is 15 the first case and −1 in the second case. Hence, the remainder when 15 is divided by 16 is 15 or −1. Î When a number N < D gives a remainder R (= N) when divided by D, it gives a negative remainder of R − D. For example, when a number gives a remainder of −2 with 23, it means that the number gives a remainder of 23 – 2 = 21 with 23. Find the remainder when 752 is divided by 2402. Answer: 752 = (74)13 = (2401)13 = (2402 – 1)13 = 2402K + (−1)13 = 2402K −1. Hence, the remainder when 752 is divided by 2402 is equal to −1 or 2402 – 1 = 2401. Remainder = 2401. When dividend is of the form an + bn or an – bn Theorem1: an + bn is divisible by a + b when n is ODD. Theorem 2: an – bn is divisible by a + b when n is EVEN. Theorem 3: an – bn is ALWAYS divisible by a – b. What is the remainder when 3444 + 4333 is divided by 5? Answer: The dividend is in the form ax + by. We need to change it into the form an + bn. 3444 + 4333 = (34)111 + (43)111. Now (34)111 + (43)111 will be divisible by 34 + 43 = 81 + 64 = 145. Since the number is divisible by 145 it will certainly be divisible by 5. Hence, the remainder is 0. What is the remainder when (5555)2222 + (2222)5555 is divided by 7? Answer: The remainders when 5555 and 2222 are divided by 7 are 4 and 3 respectively. Hence, the problem reduces to finding the remainder when (4)2222 + (3)5555 is divided by 7. Now (4)2222 + (3)5555 = (42)1111 + (35)1111 = (16)1111 + (243)1111. Now (16)1111 + (243)1111 is divisible by 16 + 243 or it is divisible by 259, which is a multiple of 7. Hence the remainder when (5555)2222 + (2222)5555 is divided by 7 is zero. 202004 + 162004 – 32004 − 1 is divisible by: (a) 317 (b) 323 (c) 253 (d) 91 Answer: 202004 + 162004 – 32004 – 1 = (202004 – 32004) + (162004 – 12004). Now 202004 – 32004 is divisible by 17 (Theorem 3) and 162004 – 12004 is divisible by 17 (Theorem 2). Hence the complete expression is divisible by 17. 202004 + 162004 – 32004 – 1 = (202004 – 12004) + (162004 – 32004). Now 202004 – 12004 is divisible by 19 (Theorem 3) and 162004 – 32004 is divisible by 19 (Theorem 2). Hence the complete expression is also divisible by 19. Hence the complete expression is divisible by 17 × 19 = 323. When f(x) = a + bx + cx2 + dx3 +... is divided by x – a http://www.totalgadha.com The remainder when f(x) = a + bx + cx2 + dx3 +.. is divided by x – a is f(a). Therefore, If f(a) = 0, (x − a) is a factor of f(x). What is the remainder when x3 + 2x2 + 5x + 3 is divided by x + 1? Answer: The remainder when the expression is divided by (x − (−1)) will be f(−1). Remainder = (−1)3 + 2(−1)2 + 5(−1) + 3 = −1 If 2x3 −3x2 + 4x + c is divisible by x – 1, find the value of c. Answer: Since the expression is divisible by x – 1, the remainder f(1) should be equal to zero ⇒ 2 – 3 + 4 + c = 0, or c = −3. Euler’s Theorem Mφ(N) If M and N are two numbers coprime to each other, i.e. HCF(M,N) = 1 and N = apbqcr ..., Re mainder[ ] = 1, where N 1 1 1 φ(N) = N(1 − )(1 − )(1 − )...and is known as Euler ' s Totient function.. φ(N) is also the number of numbers less than a b c and prime to N. Find the remainder when 537 is divided by 63. Answer: 5 and 63 are coprime to each other, therefore we can apply Euler’s theorem here. 1 3 1 7 63 = 32 × 7 ⇒ φ(63) = 63(1 − )(1 − ) = 36 Therefore, Re mainder[ 537 536 × 5 ] = Re mainder[ ]=5 63 63 Find the last three digits of 57802. Answer: Many a times (not always), the quicker way to calculate the last three digits is to calculate the remainder by 1 000. We can see that 57 and 1 000 are coprime to each other. Therefore, we can use Euler’s theorem here if it’s useful. 1 2 1 4 1 000 = 23 × 53 ⇒ φ(1000) = 1000(1 − )(1 − ) = 400 Therefore, 57400 57400 × 57400 57800 Re mainder[ ] = 1 ⇒ Re mainder[ ] = Re mainder[ ]=1 1000 1000 1000 57802 57800 × 572 ] = Re mainder[ ] = 249 ⇒ Re mainder[ 1000 1000 Hence, the last two digits of 57802 are 249. Fermat’s Theorem If p is a prime number and N is prime to p, then Np – N is divisible by p. What is the remainder when n7 – n is divided by 42? http://www.totalgadha.com Answer: Since 7 is prime, n7 – n is divisible by 7. n7 – n = n(n6 – 1) = n (n + 1)(n – 1)(n4 + n2 + 1). Now (n – 1)(n)(n + 1) is divisible by 3! = 6. Hence n7 – n is divisible by 6 x 7 = 42. Hence the remainder is 0. Fermat's Little Theorem 1 N If N in the above Euler’s theorem is a prime number, then φ(N) = N(1 − ) = N − 1 . Therefore, if M and N are MN−1 ]=1 N coprime to each other and N is a prime number, Re mainder[ Find the remainder when 5260 is divided by 31. Answer: 31 is a prime number therefore φ(N) = 30. 52 and 31 are prime to each other. Therefore, by Fermat’s theorem: 5230 5260 Re mainder[ ] = 1 ⇒ Re mainder[ ]=1 31 31 Wilson’s Theorem If P is a prime number then Re mainder[ (P − 1)! + 1 ] = 0 . In other words, (P − 1)! + 1 is divisible by P if P is P a prime number. It also means that the remainder when (P − 1)! Is divided by P is P − 1 when P is prime. Find the remainder when 40! is divided by 41. Answer: By Wilson’s theorem, we can see that 40! + 1 is divisible by 41 ⇒ Re mainder[ 40! ] = 41 − 1 = 40 41 Find the remainder when 39! is divided by 41. Answer: In the above example, we saw that the remainder when 40! is divided by 41 is 40. ⇒ 40! = 41k + 40 ⇒ 40 × 39! = 41k + 40. The R.H.S. gives remainder 40 with 41 therefore L.H.S. should also give remainder 40 with 41. L.H.S. = 40 × 39! where 40 gives remainder 40 with 41. Therefore, 39! should give remainder 1 with 41. Chinese Remainder Theorem This is a very useful result. It might take a little time to understand and master Chinese remainder theorem completely but once understood, it is an asset. If a number N = a × b, where a and b are prime to each other, i.e., hcf(a, b) = 1, and M is a number M a M b M N such that Re mainder[ ] = r1 and Re mainder[ ] = r2 then Re mainder[ ] = ar2x + br1y, where ax + by = 1 Confused? Following example will make it clear. Find the remainder when 3101 is divided by 77. http://www.totalgadha.com Answer: 77 = 11 × 7. By Fermat’s little theorem, Re mainder[ 36 310 ] = 1 AND Re mainder[ ]=1 7 11 3101 396 × 35 (36 )16 × 35 1 × 35 ] = Re mainder[ ] = Re mainder[ ] = Re mainder[ ] = 5 = r1 7 7 7 7 3101 3100 × 3 (310 )10 × 3 1× 3 Re mainder[ ] = Re mainder[ ] = Re mainder[ ] = Re mainder[ ] = 3 = r2 11 11 11 11 Re mainder[ Now we will find x and y such that 7x + 11y = 1. By observation we can find out, x = −3 and y = 2. Now we can say that Re mainder[ 3101 ] = 7 × 3 × −3 + 11 × 5 × 2 = 47 77 We can also solve this problem by Euler’s theorem and this is the method I follow most of the time. No confusion remains thereby. Find the remainder when 3101 is divided by 77. 1 7 1 ) = 60 11 360 3101 360 × 341 1 × 341 341 Re mainder[ ] = 1 ⇒ Re mainder[ ] = Re mainder[ ] = Re mainder[ ] = Re mainder[ ] 77 77 77 77 77 Answer: φ(77) = 77(1 − )(1 − Re mainder[ 34 81 ] = Re mainder[ ] = 4 77 77 ⇒ Re mainder[ = Re mainder[ 341 (34 )10 × 3 410 × 3 44 × 44 × 42 × 3 256 × 256 × 48 ] = Re mainder[ ]Re mainder[ ] = Re mainder[ ] = Re mainder[ ] 77 77 77 77 77 25 × 25 × 48 9 × 48 ] = Re mainder[ ] = 47 77 77 Find the smallest number that when divided by 7, 8 and 9 leave a remainder of 5, 4 and 3. Answer: Let N be the number. Therefore, N = 7a + 5, N = 8b + 4, N = 9c + 3. 32 Find the remainder when 3232 is divided by 9. 1 3 Answer: Notice that 32 and 9 are coprime. φ(9) = 9(1 − ) = 6 Hence by Euler’s theorem, Re mainder[ 326 ] = 1 . Since the power is 3232, we will have to simplify this 9 power in terms of 6k + r. Therefore, we need to find the remainder when 3232 is divided by 6. Re mainder[ 3232 232 (28 )4 256 × 256 × 256 × 256 256 ] = Re mainder[ ] = Re mainder[ ] = Re mainder[ ] = Re mainder[ ]= 4 6 6 6 6 6 http://www.totalgadha.com 32 Therefore, 3232 = 326k + 4 = (326)k × 324 (326 )k × 324 324 5×5×5×5 625 ] = Re mainder[ ] = Re mainder[ ] = Re mainder[ ]= 4 9 9 9 9 ⇒ Re mainder[ What will be the remainder when N = 1010 + 10100 + 101000 +........... + 1010000000000 is divided by 7? Answer: By Fermat’s Little Theorem 106 will give remainder as 1 with 7. 1010 106 × 104 104 34 Re mainder[ ] = Re mainder[ ] = Re mainder[ ] = Re mainder[ ] = 4 7 7 7 7 Similarly, all the other terms give remainder of 4 with 7. Therefore, total remainder = 4 + 4 + 4… (10 times) = 40. Remainder of 40 with 7 = 5 What is the remainder when N = 22225555 + 55552222 is divided by 7? 22226 will give remainder 1 when divided by 7. 5555 = 6K+5 ⇒ 22225555 = 22226k + 5 ⇒ Re mainder[ 22225555 22225 35 ] = Re mainder[ ] = Re mainder[ ] = 5 7 7 7 Also 55556 will give remainder 1 when divided by 7. 55552222 55552 42 ] = Re mainder[ ] = Re mainder[ ] = 2 7 7 7 55552222 = 55556k + 2 ⇒ Re mainder[ So final remainder is (5 + 2) divided by 7 = 0 Find the remainder when 8643 is divided by 132. Answer: Note that here 8 and 132 are not coprime as HCF (8, 132) = 4 and not 1. Therefore, we cannot apply Euler’s theorem directly. Re mainder[ 8643 21929 21927 ] = Re mainder[ ] = 4 × Re mainder[ ]. Now we can apply Euler ' s theorem. 132 132 33 φ(33) = 33(1 − 1 1 220 21927 27 )(1 − ) = 20 ⇒ Re mainder[ ] = 1 ⇒ Re mainder[ ] = Re mainder[ ] = 29 3 11 33 33 33 ⇒ Re al remainder = 4 × 29 = 116 To find the number of numbers that are less than or equal to a certain natural number n, and that are divisible by a certain integer http://www.totalgadha.com To find the number of numbers, less than or equal to n, and that are divisible by a certain integer p, we divide n by p. The quotient of the division gives us the number of numbers divisible by p and less than or equal to n. How many numbers less than 400 are divisible by 12? Answer: Dividing 400 by 12, we get the quotient as 33. Hence the number of numbers that are below 400 and divisible by 12 is 33. How many numbers between 1 and 400, both included, are not divisible either by 3 or 5? Answer: We first find the numbers that are divisible by 3 or 5. Dividing 400 by 3 and 5, we get the quotients as 133 and 80 respectively. Among these numbers divisible by 3 and 5, there are also numbers which are divisible both by 3 and 5 i.e. divisible by 3 x 5 = 15. We have counted these numbers twice. Dividing 400 by 15, we get the quotient as 26. Hence the number divisible by 3 or 5 = 133 + 80 – 26 = 187 Hence, the numbers not divisible by 3 or 5 are = 400 – 187 = 213. How many numbers between 1 and 1200, both included, are not divisible by any of the numbers 2, 3 and 5? Answer: as in the previous example, we first find the number of numbers divisible by 2, 3, or 5. from set theory we have n(AUBUC) = n(A) + n(B) + n(C) – n(A∩B) – n(B∩C) – n(A∩C) + n(A∩B∩C) n(2U3U5) = n(2) + n(3) + n(5) – n(6) – n(15) – n(10) + n(30) Æ n(2U3U5) = 600 + 400 + 240 – 200 – 80 – 120 + 40 = 880 Hence number of numbers not divisible by any of the numbers 2, 3, and 5 = 1200 – 880 = 320. Some Special Problems: Find the remainder when 123456789101112……..40 is divided by 36. Answer: 36 = 9 × 4. Therefore, we first find the remainders when this number is divided by 9 and 4. The remainder by 9 would be the remainder when the sum of digits is divided by 9. Sum of digits = 4 × (1 + 2 + 3 + 4 + … + 9) + 10 × (1 + 2 + 3) + 4 = 180 + 60 + 4 = 244 ⇒ remainder by 9 = 1. The remainder by 4 would be the remainder when the last two digits are divided by 4 ⇒ remainder by 4 = 0. Therefore, to find the remainder we need to find the smallest multiple of 4 that gives remainder 1 with 9. The smallest such number = 28. Therefore, remainder = 28. Find the remainder when 112123123412345…12345678 is divided by 36. Answer: 36 = 9 × 4. Therefore, we first find the remainders when this number is divided by 9 and 4. The remainder by 9 would be the remainder when the sum of digits is divided by 9. Sum of digits = 1 × 8 + 2 × 7 + 3 × 6 + … + 8 × 1 = 120 ⇒ remainder by 9 = 3. The remainder by 4 would be the remainder when the last two digits are divided by 4 ⇒ remainder by 4 = 2. The overall remainder would be the smallest number that gives remainder 3 with 9 and remainder 2 with 4. Therefore, the number would satisfy the equation 9a + 3 = 4b + 2 ⇒ 4b − 9a = 1 ⇒ (a, b) = (3, 7) and the number = 30. Therefore, remainder = 30. http://www.totalgadha.com Let n! = 1 × 2 × 3 × … × n for integer n ≥ 1. If p = 1! + (2 × 2!) + (3 × 3!) + … + (10 × 10!), then p+2 when divided by 11! leaves a remainder of (CAT 2005) 1. 10 2. 0 3. 7 4. 1 Answer: Nth term of the series = n × n! = (n + 1 − 1) × n! = (n + 1)! − n! Therefore, p = 2! − 1! + 3! − 2! + 4! − 3! + … + 11! − 10! = 11! − 1! ⇒ p + 2 = 11! + 1 ⇒ remainder by 11! = 1 Find the remainder when 1 × 2 + 2 × 3 + 3 × 4 + … + 98 × 99 + 99 × 100 is divided by 101. Answer: Nth term of the series = n × (n + 1) = n2 + n. Therefore, sum of the series ∑ (n2 + n) = n(n + 1)(2n + 1) n(n + 1) n(n + 1)(n + 2) 99 × 100 × 101 + = = 6 2 6 6 ⇒ remainder by 101 = 0. A number when divided by 8 leaves remainder 3 and quotient Q. The number when divided by 5 leaves remainder 2 and quotient Q + 8. What is the number? Answer: Let the number be N ⇒ N = 8Q + 3 and N = 5(Q + 8) + 2 = 5Q + 42 8Q + 3 = 5Q + 42 ⇒ Q = 13 ⇒ N = 107 Find the largest natural number that divides 364, 414, and 539 and leaves the same remainder in each case. Answer: Let the divisor be D and the remainder be R. Therefore, 364 = Da + R, 414 = Db + R, 539 = Dc + R Subtracting first equation from the second and the second equation from the third we get 50 = D (b − a) and 125 = D (c − b). As D is the common factor in RHS of both the equation, it should be the common factor on the LHS of both the equation. The HCF of 50 and 125 is 25. Therefore, the highest number can be 25. What is the remainder when 11111...11111  243 times is divided by 243? Answer: It can be proved that a number formed by writing any single digit 3n times will be divisible by 3n. This is left to students to check it out. How many numbers between 1 and 1000 are there such that n2 + 3n + 5 is divisible by 121? Answer: 0 values. n2 + 3n + 5 = (n - 4)(n + 7) + 33. Now, 33 is divisible by 11 but not 121. n + 7 and n - 4 are two numbers with a difference of 11, therefore either both are divisible by 11 or both are not divisible by 11. If both are divisible by 11, their product is divisible by 121 but 33 is divisible only by 11 therefore the expression is not divisible by 121. If both are not divisible by 11, the expression is again not divisible by 121. Find the remainder when 139 + 239 + 339 + 439 + ... + 1239 is divided by 39. Answer: 1p + 2p + 3p + … np is divisible by 1 + 2 + 3 + … + n if p is odd. Therefore, remainder = 0 as 1 + 2 + 3 + … + 12 = 78 which is a factor of 13. DIVISORS OF A NUMBER Divisors: http://www.totalgadha.com For a natural number N, all the numbers, including 1 and N itself, which divide N completely are called divisors of N. Example: The number 24 is divisible by 1, 2, 3, 4, 6, 8, 12, and 24. Hence all these numbers are divisors of 24. How to find the number of divisors of a number: Let us the find the number of divisors of 60. 60 = 22 × 3 × 5. Any divisors of 60 will have powers of 2 equal to either 20 or 21 or 22. Similarly, any divisor of 60 will have powers of 3 equal to either 30 or 31, and powers of 5 equal to either 50 or 51. To make a divisor of 60, we will have to choose a power of 2, a power of 3 and a power of 5. A power of 2 can be chosen in 3 ways out of 20 or 21, or 22. Similarly, a power of 3 can be chosen in 2 ways and a power of 5 can be chosen in 2 ways. Therefore, the number of divisors = 3 × 2 × 2 = 12. Notice that we have added 1 each to powers of 2, 3 and 5 and multiplied. Now for the formula: Let N be a composite number such that N = (x)a(y)b(z)c.. where x, y, z.. are prime factors. Then, the number of divisors of N = (a + 1)(b + 1)(c + 1).. Find the number of divisors of 21600. Answer: 21600 = 25 × 33 × 52 ⇒ Number of divisors = (5 + 1) × (3 + 1) × (2 + 1) = 6 × 4 × 3 = 72. How many divisors of 21600 are odd numbers? Answer: An odd number does not have a factor of 2 in it. Therefore, we will consider all the divisors having powers of 3 and 5 but not 2. Therefore, ignoring the powers of 2, the number of odd divisors = (3 + 1) × (2 + 1) = 4 × 3 = 12. How many divisors of 21600 are even numbers? Answer: Total number of divisors of 21600 = 72. Number of odd divisors of 21600 = 12. ⇒ Number of even divisors of 21600 = 72 − 12 = 60. How many divisors of 360 are not divisors of 540 and how many divisors of 540 are not divisors of 360? Answer: The best option here is to find the number of common divisors of 360 and 540. For that we find the highest common powers of all the common prime factors in 360 and 540. Now, 360 = 23 × 32 × 5 and 540 = 22 × 33 × 5. The number of common factors would be made by 22 × 32 × 5. The number of factors made by this = 3 × 3 × 2 = 18. Therefore, the two numbers will have 18 factors in common. Number of factors of 360 = 4 × 3 × 2 = 24 ⇒ Number of factors of 360 which are not factors of 540 = 24 − 18 = 6. Number of factors of 540 = 3 × 4 × 2 = 24 ⇒ Number of factors of 540 which are not factors of 360 = 24 − 18 = 6. http://www.totalgadha.com How many divisors of the number 27 × 35 × 54 have unit digit equal to 5? Answer: For unit digit equal to 5, the number has to be a multiple of 5 and it should not be a multiple of 2 otherwise the unit digit will be 0. To be a multiple of 5, the powers of 5 that it can have is 51, 52, 53 or 54. The powers of 3 can be 30, 31, 32, 33, 34 or 35. Therefore, the number of divisors which have a unit digit of 5 = 4 × 6 = 24. How many divisors of 3636 are perfect cubes? Answer: 3636 = 272372. To find the divisors which are perfect cubes, we need to take those powers of prime factors which are multiples of 3. Therefore, powers of 2 will be 20, 23, 26, 29, … 272 and similarly, powers of 3 will be 30, 33, 36, 39, … 372. Both are 25 in number. Therefore, number of divisors = 25 × 25 = 625. Reverse Operations on Divisors: Find all the numbers less than 100 which have exactly 8 divisors. Answer: To find the number of divisors of a number, we used to add 1 to powers of all the prime factors and then multiply them together. Now, given the number of divisors, we will express this number as a product and then subtract 1 from every multiplicand to obtain the powers. 8 = 2 × 2 × 2 = (1 + 1) × (1 + 1) × (1 + 1). Therefore, the number is of the form a1b1c1, where a, b and c are prime. The numbers can be 2 × 3 × 5 = 30, 2 × 3 × 7 = 42, 2 × 3 × 11 = 66, 2 × 3 × 13 = 78, 2 × 5 × 7 = 70. 8 = 4 × 2 = (3 + 1) × (1 + 1). Therefore, the number is of the form a3b, where a and b are prime. The numbers can be 23 × 3 = 24, 23 × 5 = 40, 23 × 7 = 56, 23 × 11 = 88, 33 × 2 = 54. The number can also be of the form a7, but there is no such number less than 100. Find the smallest number with 15 divisors. Answer: 15 = 3 × 5 = (2 + 1)(4 + 1) ⇒ The number is of the form a2b4, where a and b are prime. To find the smallest such number, we give the highest power to smallest rime factor, i.e. 2, and the next highest power to next smallest prime number, i.e. 3, and so on. Therefore, the smallest number = 24 × 32 = 144. Let N be a composite number such that N = (2)a(y)b(z)c.. where y, z.. are prime factors. Then, the number of even divisors of N = (a)(b + 1)(c + 1) and number of even divisors of N = (b + 1)(c + 1) How many divisors of 21600 are perfect squares? Answer: In a perfect square, all the prime factors have even powers. For example, 25 × 68 will not be a perfect square as the power of 2 is odd whereas 24 × 68 will be a perfect square because all the prime factors have even powers. 21600 = 25 × 33 × 52 therefore, all the divisors made by even powers of 2, 3 and 5 will be perfect squares. The even powers of 2 are 20, 22, 24, even powers of 3 are 30 and 32, and even powers of 5 are 50 and 52. We can select an even power of 2 in 3 ways, even power of 3 in 2 ways, and even power of 5 in 2 ways. Therefore, the number of combinations = 3 × 2 × 2 = 12. Let N be a composite number such that N = (x)a(y)b(z)c.. where x, y, z.. are prime factors. Then, the sum of divisors of N = xa+1 − 1 yb +1 − 1 zc +1 − 1 ... × × x −1 y −1 z −1 http://www.totalgadha.com What is the sum of divisors of 60? Answer: 60 = 22 × 3 × 5 ⇒ Sum of the divisors = 23 − 1 32 − 1 52 − 1 × × = 168 2 −1 3 −1 5 −1 Find the sum of even divisors of 25 × 35 × 54 Answer: All the even divisors of the number will have powers of 2 equal to one of 2, 22, 23, 24, or 25. Therefore, sum of even divisors = (2 + 22 + 23 + 24 + 25) × (1 + 3 + 32 + 33 + 34 + 35) × (1 + 5 + 52 + 53 + 54) = 2(25 − 1) 36 − 1 55 − 1 × × = 17625608 2 −1 3 −1 5 −1 Let N be a composite number such that N = (x)a(y)b(z)c.. where x, y, z.. are prime factors. Then, the product of divisors of N = (N) (a+1)(b +1)(c +1) (a+1)(b +1)(c +1) 2 = ⎛⎜ xaybzc ⎞⎟ 2 ⎝ ⎠ = xa(a +1)yb(b +1)zc(c +1) What is the product of divisors of 60? Answer: 60 = 22 × 3 × 5 ⇒ product of divisors of 60 = ( 60 ) 3×2×2 = 606 = 212 × 36 × 56 2 Let A = set of all divisors of 8100 and B = set of all divisors of 21600. What is the product of the elements of AUB? Answer: 8100 = 22 × 34 × 52 and 21600 = 25 × 33 × 52. AUB will have all the divisors of 8100 and 21600 with the common divisors written only once. Therefore, these common divisors will be multiplied only once. The common divisors will come from 22 × 33 × 52 and are 36 in number. Their product will be (22 × 33 × 52)18 = 236 × 354 × 536 Required product= 45 product of divisors of 8100 × product of divisors of 21600 (22 × 34 × 52 ) 2 × (25 × 33 × 52 )36 = = 2189 × 3144 × 581 product of common divisors 236 × 354 × 536 Let N be a composite number such that N = (x)a(y)b(z)c.. where x, y, z.. are prime factors. If N is not a perfect square, then, the number of ways N can be written as a product of two numbers = (a + 1)(b + 1)(c + 1) Number of divisors = 2 2 If N is a perfect square, then, the number of ways N can be written as a product of two numbers = (a + 1)(b + 1)(c + 1) + 1 Number of divisors + 1 = 2 2 For example, the divisors of 60 are 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, and 60. Now, 60 = 1 × 60 = 2 × 30 = 3 × 20 = 4 × 15 = 5 × 12 = 6 × 10. Therefore, divisors occur in pairs for numbers which are not perfect squares. The divisors of 36 are 1, 2, 3, 4, 6, 9, 12, 18 and 36. 36 = 1 × 36 = 2 × 18 = 3 × 12 = 4 × 9 = 6 × 6. Therefore, divisors occur in pairs except for the square root for numbers which are perfect squares. http://www.totalgadha.com How many ordered pairs of integers, (x, y) satisfy the equation xy = 110? Answer: 110 = 2 × 5 × 11. Hence, the number of divisors of 110 is = 2 × 2 × 2 = 8. Hence, the number of positive ordered pairs of x and y = 8 (as (2, 55) is not same as (55, 2)). Also, since we are asked for integers, the pair consisting of two negative integers will also suffice. Hence the total number of ordered pairs = 2 × 8 = 16. The number of ways in which a composite number can be resolve