# 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

Recall that $1000$ divides this expression if $8$ and $125$ 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 $5,1,5,1,\ldots$, while the RHS goes $1,2,3,4,5,6,7,8,1,\ldots$. The cycle length of the LHS is $2$, RHS is $8$, so the cycle length of the solution is $\operatorname{lcm}(2,8)=8$. Indeed, the $n$ that solve this congruence are exactly those such that $n \equiv 5 \pmod{8}$.

(2) $2^n \equiv n \pmod{125}$. This is extremely computationally intensive if we try to calculate all $2^1,2^2,\ldots,2^{100} \pmod{125}$, so we take a divide-and-conquer approach instead. In order for this expression to be true, $2^n \equiv n \pmod{5}$ is necessary; it shouldn't take too long for us to go through the $20$ possible LHS-RHS combinations, considering that $n$ must be odd. We only need to test $10$ values of $n$ and obtain $n \equiv 3 \pmod{20}$ or $n \equiv 17 \pmod{20}$.

With this in mind we consider $2^n \equiv n \pmod{25}$. 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's cycle length is $20$ and the RHS's cycle length is $25$, from which their least common multiple is $100$. In this step we need to test all the numbers between $1$ to $100$ that $n \equiv 3 \pmod{20}$ or $n \equiv 17 \pmod{20}$. In the case that $n \equiv 3 \pmod{20}$, the RHS goes $3,23,43,63,83$, and we need $2^n \equiv n \equiv 2^3 \pmod{25}$; 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 $2^{97}$ and $2^{83}$ modulo $125$:

Note that $2^{97}\equiv2^{-3}$; because $8\cdot47=376\equiv1\pmod{125},$ we get $2^{97} \equiv 47\pmod{125}$.

Note that $2^{83}$ is $2^{-17}=2^{-16}\cdot2^{-1}$. We have \begin{align*} 2^{-1}&\equiv63, \\ 2^{-2}&\equiv63^2=3969\equiv-31, \\ 2^{-4}&\equiv(-31)^2=961\equiv-39, \\ 2^{-8}&\equiv1521\equiv21, \\ 2^{-16}&\equiv441, \\ 2^{-17}&\equiv63\cdot441\equiv7\cdot(-31)=-217\equiv33. \end{align*} This time, LHS cycle is $100$, RHS cycle is $125$, so we need to figure out $n$ modulo $500$. It should be $n \equiv 283,297 \pmod{500}$.

Put everything together. By the second subcondition, the only candidates less than $1000$ are $283,297,783,797$. Apply the first subcondition, $n=\boxed{797}$ is the desired answer.

~Ross Gao (Solution)

~MRENTHUSIASM (Minor Reformatting)

## 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 system of 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}, \hspace{15mm} &(1) \\ 2^n &\equiv n \pmod{125}. &(2) \end{align*} We solve each congruence separately:

1. For $(1),$ quick inspections produce that $5^1,5^2,5^3,5^4,\ldots$ are congruent to $5,1,5,1,\ldots$ 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, $\boldsymbol{n=8r+5}$ for some nonnegative integer $\boldsymbol{r.}$

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} \\ \Biggl[\binom{r}{0}5^0+\binom{r}{1}5^1+\binom{r}{2}5^2+\phantom{ }\underbrace{\binom{r}{3}5^3+\cdots+\binom{r}{r}5^r}_{0\pmod{125}}\phantom{ }\Biggr]\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+2r+27&\equiv0\phantom{r+5}\pmod{125}. \hspace{15mm} (*) \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, $\boldsymbol{r=5s+4}$ for some nonnegative integer $\boldsymbol{s.}$

Substituting this back into $(*),$ we get \begin{align*} 25(5s+4)^2+2(5s+4)+27&\equiv0\pmod{125} \\ 625s^2+1010s+435&\equiv0\pmod{125} \\ 10s+60&\equiv0\pmod{125} \\ 10(s+6)&\equiv0\pmod{125}. \end{align*} As $10(s+6)$ is a multiple of $125,$ it has at least three factors of $5.$ Since $10$ contributes one factor, $s+6$ contributes at least two factors, or $s+6$ must be a multiple of $25.$ Therefore, the least such nonnegative integer $s$ is $19.$

Finally, combining the two results from above (bolded) generates the least such positive integer $n$ at $s=19:$ \begin{align*} n&=8r+5 \\ &=8(5s+4)+5 \\ &=40s+37 \\ &=\boxed{797}. \end{align*} ~MRENTHUSIASM (inspired by Math Jams's 2021 AIME II Discussion)

## Solution 4 (Minimal Computation)

$$5^n \equiv n \pmod{8}$$ Note that $5^n \equiv 5,1,5,1,...$ and $n \equiv 0,..,7$ so $n$ is periodic every $[2,8]=8$. Easy to check that only $n\equiv 5 \pmod{8}$ satisfy. Let $n=8a+5$. Note that by binomial theorem, $$2^{8a+5}=32\cdot 2^{8a} \equiv 7(1+15)^{2a} \equiv 7(1+30a)\pmod{25}$$ So we have $7(1+30a) \equiv 8a+5\pmod{25} \implies 202a \equiv 23 \implies 2a \equiv 23+25 \implies a \equiv 24 \pmod{25}$. Combining $a\equiv 24\pmod{25}$ with $n \equiv 5 \pmod{8}$ gives that $n$ is in the form of $197+200k$, $n=197,397,597,797$. Note that since $\phi(125)=100$ $$2^{197+200k} \equiv 2^{197} \equiv 2^{-3} \equiv \underbrace{47}_{\text{Extended Euclidean Algorithm}}\pmod{125}$$ Easy to check that only $\boxed{797}\equiv 47\pmod{125},{1797}\equiv 47\pmod{125},{2797}\equiv 47\pmod{125},...$

~Afo

## Solution 5 (STEP by step transform)

1. The desired $n$ has the form $n = m + 20l + 100p,$ where $m, l, p$ are integers and $m < 20.$

Really: $$(2^{m+ 20} – (m + 20)) – (2^m – m) = (2^{m+ 20} – 2^m) – 20.$$ The first term is a multiple of $100$(Claim).

We denote step an increase in $m$ by 20. At each step, the divisibility by $10$ is preserved, the number of tens is reduced by $2$.

$$(2^{m+ 100} – (m + 100)) – (2^m – m) = (2^{m+ 100} – 2^m) – 100.$$ We denote STEP an increase in $m$ by $100.$ At each STEP, the first term is a multiple of $1000,$ which means that at each STEP the divisibility by $100$ is preserved, the number of hundreds decreases by $1.$

2. If the expression $2^n + 5^n – n$ is a multiple of $1000,$ the number $n$ is odd ($2^n$ is even, $5^n$ is odd), which means that $5^n$ ends in $125.$ Therefore, the number $2^n – n$ ends in $875.$

3. $2^3 – 3 = 8 – 3 = 5.$ The tens digit is even $(0),$ it cannot be transformed into $7$ in several steps, since at each step this digit changes by $2.$

$17 = 20 – 3,$ so $2^{17} + 2^3$ is a multiple of $10,$ $(2^{17} – 17) + (2^3 – 3) = 2^{17} + 2^3 – 20$ is a multiple of $10.$ $$2^{17} \pmod {100} = ((2^{10} \pmod {100}) \cdot (2^7 \pmod {100})) \pmod {100} = (24 \cdot 28) \pmod {100} = 72.$$ $$(2^{17} – 17 ) \pmod {100} = 55.$$ The tens digit $5$ is odd, subtracting $2$ at each step in $4$ steps of $20$ will turn it into $7.$ So $$(2^{97} – 97 ) \pmod {100} = 75.$$ $$2^{97} \pmod {1000} = ((2^{49} \pmod {1000}) \cdot 2^7) \pmod {1000} = 672.$$ $$(2^{97} – 97 ) \pmod {1000} = 575.$$ We transform $5$ into $8$ in $7$ STEPS, so $$(2^{797} – 797 ) \pmod {1000} = 875.$$ $$(5^{797} + 2^{797} – 797 ) \pmod {1000} = 0.$$ Note, that for $n = 797 + 1000k, k$ is an integer, the expression $2^n + 5^n – n$ is a multiple of $1000.$

Claim

The numbers $2^{n+2\cdot 5^k}+2^n=2^n(2^{2\cdot 5^k}+1)$ and $2^{n+4\cdot 5^k}-2^n=2^n(2^{4\cdot 5^k}-1)$

are a multiple of $10^{k+1}$ for any $n > k$, where $n,k$ are an integer.

Proof

First, if $m \pmod{10} \equiv 4,$ then $(m^4 – m^3 + m^2 – m + 1) \pmod{10} \equiv 6 – 4 + 6 – 4 + 1 = 5.$

$2^{4n+2}\pmod{10} \equiv 4.$ So, if the number $2^{4n+2} + 1$ is a multiple of $5^k$, then $2^{20n+10} + 1$ is a multiple of $5^{k +1}.$

Really, denote $m = 2^{4n+2},$ then, $2^{20n+10} + 1 = m^5 + 1 = (m + 1)\cdot (m^4 – m^3 + m^2 – m + 1).$

The first factor is a multiple of $5^k$, the second factor is a multiple of $5.$ Their product is a multiple of $5^{k +1}.$

For odd $n,$ using Newton's binomial for $4^n$, we get $2^{2n} + 1$ is a multiple of $5.$ \begin{align*} 2^{2n} + 1 = 4^n + 1 = (5 - 1)^n + 1 = 5^n -n\cdot 5^{n-1} +...+5n\end{align*} If $n = 5^k,$ then $n > k,$ each term is a multiple of $5n$.

Therefore, the number $4^{5^k}+1$ is a multiple of $5^{k+1}$ and $2^{k+1} (2^{2\cdot 5^k}+1)$ is a multiple of $10^{k+1}$.

The difference $$2^{n+4\cdot 5^k}-2^n=2^n(2^{4\cdot 5^k}-1) = 2^n(2^{2\cdot 5^k}-1)(2^{2\cdot 5^k}+1)$$ is a multiple of $10^{k+1}.$

## Note

If you are wondering, $\frac{2^{797}+5^{797}-797}{1000}$ is equal to the following $555$-digit number:

$119{,}975{,}745{,}111{,}650{,}476{,}385{,}411{,}555{,}010{,}245{,}228{,}283{,}196{,}935{,}699{,}128{,}663{,}834{,}986{,}755{,}809{,}416{,}868{,}468{,}175{,}224{,}221{,}677{,} \\ 450{,}271{,}793{,}186{,}899{,}399{,}171{,}810{,}240{,}468{,}879{,}630{,}691{,}320{,}210{,}517{,}618{,}332{,}910{,}064{,}430{,}017{,}422{,}410{,}572{,}497{,}874{,}327{,}116{,} \\ 662{,}273{,}049{,}144{,}209{,}532{,}072{,}513{,}248{,}821{,}826{,}125{,}454{,}909{,}131{,}367{,}242{,}561{,}087{,}064{,}439{,}357{,}641{,}814{,}942{,}784{,}478{,}465{,}638{,} \\ 163{,}997{,}895{,}630{,}266{,}780{,}260{,}171{,}070{,}180{,}244{,}384{,}687{,}160{,}174{,}154{,}618{,}815{,}254{,}719{,}975{,}476{,}350{,}789{,}415{,}969{,}908{,}281{,}131{,} \\ 044{,}034{,}581{,}864{,}573{,}596{,}962{,}567{,}993{,}948{,}922{,}431{,}587{,}652{,}735{,}290{,}158{,}293{,}666{,}986{,}616{,}531{,}419{,}688{,}663{,}485{,}548{,}648{,}995{,} \\ 509{,}247{,}893{,}796{,}528{,}146{,}109{,}144{,}143{,}744{,}650{,}882{,}271{,}050{,}771{,}309{,}301{,}498{,}467{,}898{,}226{,}649{,}089{,}924{,}520{,}539{,}820{,}344{,}035{,} \\ 886{,}481{,}987{,}499{,}589{,}845{,}734{,}228{,}834{,}491{,}187.$

## Video Solution by Interstigation

Did not use any advanced method, easily understandable:

~Interstigation