Find the Min Sum of $m$ and $n$ for $(4^m+4^n)\ mod\ 100=0$

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SUMMARY

The discussion focuses on finding the minimum sum of integers \(m\) and \(n\) such that \((4^m + 4^n) \mod 100 = 0\) with the condition \(m > n\). The solution reveals that \(4^5 \equiv -1 \pmod{25}\), leading to the conclusion that \(4^6 + 4^1 \equiv 0 \pmod{100}\). Therefore, the minimum value of \(m+n\) is determined to be \(7\) (where \(m=6\) and \(n=1\)).

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  • Understanding of modular arithmetic, specifically modulo 100 and modulo 25.
  • Familiarity with powers of 4 and their properties.
  • Knowledge of basic number theory concepts, including congruences.
  • Ability to manipulate algebraic expressions involving exponents.
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This discussion is beneficial for mathematicians, students studying number theory, and anyone interested in solving modular equations and understanding their properties.

Albert1
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if $(4^m+4^n)$ mod 100=0
(here $m,n\in N \,\, and \,\,m>n$)

please find:$min(m+n)$
 
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[sp]$4^m+4^n = 4^n(4^{m-n}+1)$. If $4^{m-n} = -1\pmod{25}$ then $4^n(4^{m-n}+1) = 0 \pmod{100}.$ The first few powers of $4$ are
$4$
$16$
$64$
$256$
$1024 = 4^5$.​
Since $4^5 = 24 = -1\pmod{25}$, it follows that $4^6 + 4^1 = 0\pmod{100}$ (and in fact $4^6+4^1 = 4096 + 4 = 4100$). So the minimum value of $m+n$ is $6+1=7.$[/sp]
 
Opalg said:
[sp]$4^m+4^n = 4^n(4^{m-n}+1)$. If $4^{m-n} = -1\pmod{25}$ then $4^n(4^{m-n}+1) = 0 \pmod{100}.$ The first few powers of $4$ are
$4$
$16$
$64$
$256$
$1024 = 4^5$.​
Since $4^5 = 24 = -1\pmod{25}$, it follows that $4^6 + 4^1 = 0\pmod{100}$ (and in fact $4^6+4^1 = 4096 + 4 = 4100$). So the minimum value of $m+n$ is $6+1=7.$[/sp]
very good , the answer is correct !
 

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