# Difference between revisions of "2021 AIME II Problems/Problem 13"

## Problem

Find the least positive integer $n$ for which $2^n + 5^n - n$ is a multiple of $1000$.

## Solution 1

1000 divides this expression iff 8, 125 both divide it. It should be fairly obvious that $n \geq 3$; so we may break up the initial condition into two sub-conditions.

(1) $5^n \equiv n \pmod{8}$. Notice that the square of any odd integer is 1 modulo 8 (proof by plugging in $1^2$, $3^2$, $5^2$, $7^2$ into modulo 8), so the LHS of this expression goes "5,1,5,1, ...", while the RHS goes "1,2,3,4,5,6,7,8,1, ...". The cycle length of the LHS is 2, RHS is 8, so the cycle length of the solution is $lcm(2,8)=8$. Indeed, the n that solve this equation are exactly those such that $n \equiv 5 \pmod{8}$.

(2) $2^n \equiv n \pmod{125}$. This is extremely computationally intensive if you try to calculate all $2^1,2^2, ..., 2^{100} \pmod{125}$, so instead, we take a divide-and-conquer approach. In order for this expression to be true $2^n \equiv n \pmod{5}$ is necessary; it shouldn't take too long for you to go through the 20 possible LHS-RHS combinations and convince yourself that $n \equiv 3 \pmod{20}$ or $n \equiv 17 \pmod{20}$.

With this in mind we consider $2^n \equiv n$ modulo 25. By the generalized Fermat's little theorem, $2^{20} \equiv 1 \pmod{25}$, but we already have n modulo 20! Our calculation is greatly simplified. The LHS cycle length is 20, RHS cycle length is 25, the lcm is 100, in this step we need to test all the numbers between 1 to 100 that $n \equiv 3 \pmod{20}$ or $n \equiv 17 \pmod{17}$. In the case that $n \equiv 3 \pmod{20}$, $2^n \equiv 2^3 \equiv 8$, and RHS goes "3,23,43,63,83"; clearly $n \equiv 83 \pmod{100}$. In the case that $n \equiv 17 \pmod{20}$, by a similar argument, $n \equiv 97 \pmod{100}$.

In the final step, we need to calculate $2^{97}, 2^{83}$ modulo 125. The former is simply $2^{-3}$; because $8*47=376=1$ modulo 125, $2^{97} \equiv 47$. $2^{83}$ is $2^{-17}=2^{-16}*2^{-1}$. \begin{align*} 2^{-1}&=63, \\ 2^{-2}&=63^2=3969=-31, \\ 2^{-4}&=(-31)^2=961=-39, \\ 2^{-8}&=1521=21, \\ 2^{-16}&=441, \\ 2^{-17}&=63*441=7*{-31}=-217=33. \end{align*} This time, LHS cycle is 100, RHS cycle is 125, so we need to figure out what is n modulo 500. It should be $n \equiv 283,297 \pmod{500}$.

Put everything together. By the second subcondition, the only candidates < 100 are $283,297,782,797$. Apply the first subcondition, n=797 is the desired answer.

-Ross Gao

## Solution 2

We have that $2^n + 5^n \equiv n\pmod{1000}$, or $2^n + 5^n \equiv n \pmod{8}$ and $2^n + 5^n \equiv n \pmod{125}$ by CRT. It is easy to check $n < 3$ don't work, so we have that $n \geq 3$. Then, $2^n \equiv 0 \pmod{8}$ and $5^n \equiv 0 \pmod{125}$, so we just have $5^n \equiv n \pmod{8}$ and $2^n \equiv n \pmod{125}$. Let us consider both of these congruences separately.

First, we look at $5^n \equiv n \pmod{8}$. By Euler's Totient Theorem (ETT), we have $5^4 \equiv 1 \pmod{8}$, so $5^5 \equiv 5 \pmod{8}$. On the RHS of the congruence, the possible values of $n$ are all nonnegative integers less than $8$ and on the RHS the only possible values are $5$ and $1$. However, for $5^n$ to be $1 \pmod{8}$ we must have $n \equiv 0 \pmod{4}$, a contradiction. So, the only possible values of $n$ are when $n \equiv 5 \pmod{8} \implies n = 8k+5$.

Now we look at $2^n \equiv n \pmod{125}$. Plugging in $n = 8k+5$, we get $2^{8k+5} \equiv 8k+5 \pmod{125} \implies 2^{8k} \cdot 32 \equiv 8k+5 \pmod{125}$. Note, for $2^n \equiv n\pmod{125}$ to be satisfied, we must have $2^n \equiv n \pmod{5}$ and $2^n \equiv n\pmod{25}$. Since $2^{8k} \equiv 1\pmod{5}$ as $8k \equiv 0\pmod{4}$, we have $2 \equiv -2k \pmod{5} \implies k = 5m-1$. Then, $n = 8(5m-1) + 5 = 40m-3$. Now, we get $2^{40m-3} \equiv 40m-3 \pmod{125}$. Using the fact that $2^n \equiv n\pmod{25}$, we get $2^{-3} \equiv 15m-3 \pmod{25}$. The inverse of $2$ modulo $25$ is obviously $13$, so $2^{-3} \equiv 13^3 \equiv 22 \pmod{25}$, so $15m \equiv 0 \pmod{25} \implies m = 5s$. Plugging in $m = 5s$, we get $n = 200s - 3$.

Now, we are finally ready to plug $n$ into the congruence modulo $125$. Plugging in, we get $2^{200s-3} \equiv 200s - 3 \pmod{125}$. By ETT, we get $2^{100} \equiv 1 \pmod{125}$, so $2^{200s- 3} \equiv 2^{-3} \equiv 47 \pmod{125}$. Then, $200s \equiv 50 \pmod{125} \implies s \equiv 4 \pmod{5} \implies s = 5y+4$. Plugging this in, we get $n = 200(5y+4) - 3 = 1000y+797$, implying the smallest value of $n$ is simply $\boxed{797}$. ~rocketsri

## Solution 3 (Chinese Remainder Theorem and Binomial Theorem)

We wish to find the least positive integer $n$ for which $2^n+5^n-n\equiv0\pmod{1000}.$ Rearranging gives $$2^n+5^n\equiv n\pmod{1000}.$$ Applying the Chinese Remainder Theorem, we get the following systems of linear congruences: \begin{align*} 2^n+5^n &\equiv n \pmod{8}, \\ 2^n+5^n &\equiv n \pmod{125}. \end{align*} It is clear that $n\geq3,$ from which we simplify to \begin{align} 5^n &\equiv n \pmod{8}, \\ 2^n &\equiv n \pmod{125}. \end{align}

1. For $(1),$ quick inspections produce that $5^1,5^2,5^3,5^4,\cdots$ are congruent to $5,1,5,1,\cdots$ modulo $8,$ respectively. More generally, $5^n \equiv 5 \pmod{8}$ if $n$ is odd, and $5^n \equiv 1 \pmod{8}$ if $n$ is even. As $5^n$ is always odd (so is $n$), we must have $n\equiv5\pmod{8}.$ That is, $n=8r+5$ for some integer $r.$
2. For $(2),$ we substitute the result from $(1)$ and simplify:
3. \begin{align*} 2^{8r+5}&\equiv8r+5\pmod{125} \\ \left(2^8\right)^r\cdot2^5&\equiv8r+5\pmod{125} \\ 256^r\cdot32&\equiv8r+5\pmod{125} \\ 6^r\cdot32&\equiv8r+5\pmod{125}. \end{align*} Note that $5^3=125$ and $6=5+1,$ so we apply the Binomial Theorem to the left side: \begin{align*} (5+1)^r\cdot32&\equiv8r+5\pmod{125} \\ \left[\binom{r}{0}5^0+\binom{r}{1}5^1+\binom{r}{2}5^2+\underbrace{\binom{r}{3}5^3+\cdots+\binom{r}{r}5^r}_{0\pmod{125}}\right]\cdot32&\equiv8r+5\pmod{125} \\ \left[1+5r+\frac{25r(r-1)}{2}\right]\cdot32&\equiv8r+5\pmod{125} \\ 32+160r+400r(r-1)&\equiv8r+5\pmod{125} \\ 32+35r+25r(r-1)&\equiv8r+5\pmod{125} \\ 25r^2+10r+32&\equiv8r+5\pmod{125} \\ 25r^2+2r+27&\equiv0\pmod{125}. \end{align*} Since $125\equiv0\pmod{5},$ it follows that \begin{align*} 25r^2+2r+27&\equiv0\pmod{5} \\ 2r+2&\equiv0\pmod{5} \\ r&\equiv4\pmod{5}. \end{align*} That is, $n=5s+4$ for some integer $s.$

SOLUTION IN PROGRESS. WILL FINISH WITHIN TODAY. A MILLION THANKS FOR NOT EDITING.

~MRENTHUSIASM (inspired by Math Jams's 2021 AIME II Discussion)