New answers tagged totient-function
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Euler's Totient function
I have got the answer for this question. Thank you for the clarification in the comments. I have been trying to check for the numbers of the form $X = 2^a.5^b$ and $X = 2^a.5^b.p_1.p_2...p_k$ but I ...
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Show that no positive integer ending in $133$ can be expressed in the form $3^a7^b$, where $a$ and $b$ are non-negative integers.
Just to add some details to the nice answers already posted, the subgroup of $(\mathbb{Z}/1000)^{\times}$ generated by $3$ and $7$ has order 200, so it is of index $2$ ( since $|(\mathbb{Z}/1000)^{\...
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Show that no positive integer ending in $133$ can be expressed in the form $3^a7^b$, where $a$ and $b$ are non-negative integers.
If we look at the group generated by 3 modulo 100.
3,9,27,81,43,29, etc.
There are 20 members to this group. Every element has an even number in the 10s digit. 7 is a member of the group.
All numbers ...
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Show that no positive integer ending in $133$ can be expressed in the form $3^a7^b$, where $a$ and $b$ are non-negative integers.
Proof$_{\:\!1}$ $\!\bmod 20\!:\, 7\equiv 3^{-1}$ so $\,3^a7^b\!\equiv 3^{a-b}\!\in S\equiv \{1,3,9,7\}\,$ but $\,133\!\equiv\! 13\not\in S$ $\,\ \bf\small QED$
Proof$_{\:\!2}$ $ \bmod 4\!:\, \ \ \ \...
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Given an $n$ such that for all $m \neq n$, then $\phi(n) \neq \phi(m)$, prove that $4 \mid n$
Answering for completion :
If $n$ is not a multiple of $4$, then $n$ is either odd or an odd multiple of $2$.
If $n=2k$ with $k$ odd then $\phi(n) = \phi(2k) = \phi(2)\phi(k) = \phi(k)$.
If $n$ is odd ...
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Show that no positive integer ending in $133$ can be expressed in the form $3^a7^b$, where $a$ and $b$ are non-negative integers.
The remaining other possibility is that the number, call it $n$, has the form $3^a7^b$.
If we know the three last digits of a natural number, we also know its residue class modulo $8$. As $133\equiv5\...
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Show that no positive integer ending in $133$ can be expressed in the form $3^a7^b$, where $a$ and $b$ are non-negative integers.
Not much case analysis is actually necessary here.
Say $n$ ends in $133$. It certainly isn't divisible by $2$ or $5$; so we need to show it can't be of the form $3^a 7^b$.
You can do this by looking ...
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Show that no positive integer ending in $133$ can be expressed in the form $3^a7^b$, where $a$ and $b$ are non-negative integers.
A number with no prime factors greater than $7$ must have only prime factors in $\{2,3,5,7\}$. Such a number takes the form $2^a3^b5^c7^d$. If $n\equiv 133\bmod 1000$ takes this form, it must have no $...
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