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

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==Solution 4 (Step by step transform)==
 
==Solution 4 (Step by step transform)==
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<pre style="color: blue">
 
Lemma
 
Lemma
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</pre>
 
The difference <cmath>2^{n+4\cdot 5^k}-2^n=2^n(2^{4\cdot 5^k}-1)</cmath> is a multiple of <math>10^{k+1}</math> for any <math>n > k</math>, where <math>n,k</math> are an integer.
 
The difference <cmath>2^{n+4\cdot 5^k}-2^n=2^n(2^{4\cdot 5^k}-1)</cmath> is a multiple of <math>10^{k+1}</math> for any <math>n > k</math>, where <math>n,k</math> are an integer.
  
 
The sum  <cmath>2^{n+2\cdot 5^k}+2^n=2^n(2^{2\cdot 5^k}+1)</cmath> is a multiple of <math>10^{k+1}</math> for any <math>n > k</math>, where <math>n,k</math> are an integer.
 
The sum  <cmath>2^{n+2\cdot 5^k}+2^n=2^n(2^{2\cdot 5^k}+1)</cmath> is a multiple of <math>10^{k+1}</math> for any <math>n > k</math>, where <math>n,k</math> are an integer.
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<pre style="color: blue">
 +
Proof:
 +
</pre>
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If <math>m \mod 10 = 4,</math> then  <math>(m^4 – m^3 + m^2 – m + 1) \mod 10 = 6 – 4 + 6 – 4 + 1 = 5.</math>
  
 +
If the number  <math>2^{4n+2}  + 1</math> is a multiple of <math>5^k</math>, then <math>2^{20n+10} + 1</math> is a multiple of <math>5^{k +1}.</math>
  
 +
Denote <math>m =  2^{4n+2},</math> then,  <math>2^{20n+10} + 1 =  m^5 + 1 = (m + 1)\cdot (m^4 – m^3 + m^2 – m + 1).</math>
  
 +
The first factor is a multiple of <math>5^k</math>, the second factor is a multiple of <math>5.</math>
 +
Their product is a multiple of  <math>5^{k +1}.</math>
  
−
 
+
For odd <math>n,</math> using Newton's binomial, we get:
  
 
==See Also==
 
==See Also==
 
{{AIME box|year=2021|n=II|num-b=12|num-a=14}}
 
{{AIME box|year=2021|n=II|num-b=12|num-a=14}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 01:37, 30 May 2022

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 iff $8$ 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 $\boxed{797}\equiv 47\pmod{125}$

~Afo

Solution 4 (Step by step transform)

Lemma

The difference \[2^{n+4\cdot 5^k}-2^n=2^n(2^{4\cdot 5^k}-1)\] is a multiple of $10^{k+1}$ for any $n > k$, where $n,k$ are an integer.

The sum \[2^{n+2\cdot 5^k}+2^n=2^n(2^{2\cdot 5^k}+1)\] is a multiple of $10^{k+1}$ for any $n > k$, where $n,k$ are an integer. 

Proof:

If $m \mod 10 = 4,$ then $(m^4 – m^3 + m^2 – m + 1) \mod 10 = 6 – 4 + 6 – 4 + 1 = 5.$

If the number $2^{4n+2} + 1$ is a multiple of $5^k$, then $2^{20n+10} + 1$ is a multiple of $5^{k +1}.$

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, we get:

See Also

2021 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 12 		Followed by
Problem 14
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
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