# is $7^{101} + 18^{101}$ divisible by $25$?

I am not able to find a solution for this question. I am thinking in the lines of taking out some common element like $(7\cdot 7^{100}) + (18\cdot18^{100})$ but couldn't go anywhere further.

• $18\equiv (-7)$ mod $25$ and$101$ is odd – Piquito Jan 6 '16 at 19:13
• @Piquito which is essentially CuddlyCuttlefish's answer. – user236182 Jan 6 '16 at 19:22
• Do you mean I have copied CuddlyCuttlefish's answer? If yes, you are wrong, believed me, please. – Piquito Jan 7 '16 at 19:45

$$7^{101} + 18^{101} = 7^{101} + (-7)^{101} = 7^{101} - 7^{101} = 0 \pmod {25}$$

• So I'm on the Android app, and what I see is an answer with 12 votes. The entire content of the answer is [Math Processing Error]. This is a known bug, or at least I reported it months ago... – Matt Samuel Jan 7 '16 at 0:26
• The answer is that 101 is an odd exponent, and 18 = -7 mod 25. I've noticed this problem only recently, but I think it is a MathJax problem, rather than something MSE/SE can fix. – TokenToucan Jan 7 '16 at 0:45

If $n\in\mathbb Z_{\ge 3}$ is odd, then $$a^n+b^n=(a+b)\left(a^{n-1}-a^{n-2}b\pm \cdots+b^{n-1}\right)$$

Therefore $$7^{101}+18^{101}=25\left(7^{100}-7^{99}\cdot 18\pm\cdots +18^{100}\right)$$

Sure, since $\varphi(25)=\varphi(5^2)=4\cdot 5=20$, we have: $$7^{101}+18^{101}\equiv 7^1+18^1\equiv 25 \equiv 0\pmod{25}.$$

You can write down $7^{101}+18^{101}=7\cdot7^{100}+18\cdot18^{100}=7\cdot49^{50}+18\cdot324^{50}=7\cdot(50-1)^{50}+18\cdot(325-1)^{50}$. Using Newton's Binomial, you can conclude that the last sum is a $7\cdot$(multiple of 25$+(-1)^{50})+18\cdot($multiple of 25$+(-1)^{50})$. Thus, it is a multiple of 25 $+7+18$, which is also a multiple of 25.

You might want to check Euler's theorem:

$a^{\phi(n)} = 1 \mod n$ for each $a$ which is coprime with n.

$\phi(25) = \phi(5^2) = 5^2-5=20$

So $7^{101} = 7^{100}*7 = (7^5)^{20}*7=7 \mod 25$

In the same way $18^{101} = 18 \mod 25$, so their sum is $17+8=25=0 \mod 25$, which means that the given expression is divisible by 25.

We can use lifting the exponent lemma. Let $v_5(n)$ be the exponent of the maximum power of $5$ dividing $n$.

$v_5(7^ {101}+18^{101})=v_5(25)+v_5(101)=2+0=2$.

So it is a multiple of $25$.

• It is a multiple of $25$. What you have proved is both $5^2\mid 7^{101}+18^{101}$ and $5^3\nmid 7^{101}+18^{101}$, so you've proved a stronger result. – user236182 Jan 6 '16 at 19:02
• yes, thanks, although this solution sucks for $7$ and $18$ lol, I thought it was $7$ and $8$, in that case using $LTE$ or Hansen's lemma actually makes sense – Jorge Fernández Hidalgo Jan 6 '16 at 19:03

Yes, by the binomial theorem, $18^{101}=(25-7)^{101}=25a-7^{101}$.

Since 7 and 18 are coprime to 25, and the number of elements of the group of units of the ring Z/25Z is 20, 7^101 is congruent to 7 and 18^101 is congruent to 18 modulo 25 by Lagrange's theorem. This is essentially using Euler's extension of Fermat's little theorem, in the terminology of modern algebra.