2014 AIME I Problems/Problem 8

Problem 8

The positive integers $N$ and $N^2$ both end in the same sequence of four digits $abcd$ when written in base $10$, where digit a is not zero. Find the three-digit number $abc$.


Solution 1 (similar to Solution 3)

We have that $N^2 - N = N(N - 1)\equiv 0\mod{10000}$

Thus, $N(N-1)$ must be divisible by both $5^4$ and $2^4$. Note, however, that if either $N$ or $N-1$ has both a $5$ and a $2$ in its factorization, the other must end in either $1$ or $9$, which is impossible for a number that is divisible by either $2$ or $5$. Thus, one of them is divisible by $2^4 = 16$, and the other is divisible by $5^4 = 625$. Noting that $625 \equiv 1\mod{16}$, we see that $625$ would work for $N$, except the thousands digit is $0$. The other possibility is that $N$ is a multiple of $16$ and $N-1$ is a multiple of $625$. In order for this to happen, \[N-1 \equiv -1 \pmod {16}.\] Since $625 \equiv 1 \pmod{16}$, we know that $15 \cdot 625 = 9375 \equiv 15 \equiv -1 \mod{16}$. Thus, $N-1 = 9375$, so $N = 9376$, and our answer is $\boxed{937}$.

Solution 2 (bashing)

let $N= 10000t+1000a+100b+10c+d$ for positive integer values $t,a,b,c,d$. When we square $N$ we get that \begin{align*} N^2 &=(10000t+1000a+100b+10c+d)^2\\ &=10^8t^2+10^6a^2+10^4b^2+10^2c^2+d^2+2(10^7ta+10^6tb+10^5tc+10^4td+10^5ab+10^4ac+10^3bc+10^ad+10^2bd+10cd) \end{align*}

However, we don't have to deal with this whole expression but only with its last 4 digits so it is suffices to consider only: \[2000ad+2000bc+100c^2+200bd+20cd+d^2.\] Now we need to compare each decimal digit with $1000a+100b+10c+d$ and see whether the digits are congruent in base 10. we first consider the ones digits:

$d^2\equiv d \pmod{10}.$

This can happen for only 3 values : 1, 5 and 6.

We can try to solve each case:

  • Case 1 $(d=1)$

Considering the tenths place, we have that:

$20cd=20c\equiv 10c \pmod {100}$ so $c= 0$.

Considering the hundreds place we have that

$200bd+100c^2= 200b \equiv 100b \pmod{1000}$ so again $b=0$

now considering the thousands place we have that

$2000ad+2000bc = 2000a \equiv 1000a \pmod {10000}$ so we get $a=0$ but $a$ cannot be equal to $0$ so we consider $d=5.$

  • Case 2 $(d=5)$

considering the tenths place we have that:

$20cd+20=100c+20\equiv 20 \equiv 10c \mod  {100}$ ( the extra $20$ is carried from $d^2$ which is equal to $25$) so $c=2$

considering the hundreds place we have that

$200bd+100c^2+100c= 1000b+600 \equiv600\equiv 100b \pmod{1000}$ ( the extra $100c$ is carried from the tenths place) so $b=6$

now considering the thousands place we have that

$2000ad+2000bc +1000b= 10000a+24000+ 6000\equiv0\equiv 1000a \pmod {10000}$ ( the extra $1000b$ is carried from the hundreds place) so a is equal 0 again

  • Case 3$(d=6)$

considering the tenths place we have that:

$20cd+30=120c+30\equiv 30+20c \equiv 10c \pmod {100}$ ( the extra $20$ is carried from $d^2$ which is equal to $25$) if $c=7$ then we have

$30+20 \cdot 7 \equiv 70\equiv7 \cdot 10 \pmod{100}$

so $c=7$

considering the hundreds place we have that

$200bd+100c^2+100c+100= 1200b+4900+800 \equiv200b+700\equiv 100b \pmod{1000}$ ( the extra $100c+100$ is carried from the tenths place)

if $b=3$ then we have

$700+200 \cdot 3 \equiv 300\equiv3 \cdot 100 \pmod {1000}$

so $b=3$

now considering the thousands place we have that

$2000ad+2000bc +1000b+5000+1000= 12000a+42000+ 3000+6000\equiv0\equiv 2000a+1000\equiv 1000a \pmod {10000}$ ( the extra $1000b+6000$ is carried from the hundreds place)

if $a=9$ then we have

$2000 \cdot 9+1000 \equiv 9000\equiv9 \cdot 1000 \pmod {1000}$

so $a=9$

so we have that the last 4 digits of $N$ are $9376$ and $abc$ is equal to $\boxed{937}$

Solution 3 (general)

By the Chinese Remainder Theorem, the equation $N(N-1)\equiv 0\pmod{10000}$ is equivalent to the two equations: \begin{align*} N(N-1)&\equiv 0\pmod{16},\\ N(N-1)&\equiv 0\pmod{625}. \end{align*} Since $N$ and $N-1$ are coprime, the only solutions are when $(N\mod{16},N\mod{625})\in\{(0,0),(0,1),(1,0),(1,1)\}$.

Let \[\varphi:\mathbb Z/10000\mathbb Z\to\mathbb Z/16\mathbb Z\times\mathbb Z/625\mathbb Z,\] \[x\mapsto (x\mod{16},x\mod{625}).\] The statement of the Chinese Remainder theorem is that $\varphi$ is an isomorphism between the two rings. In this language, the solutions are $\varphi^{-1}(0,0)$, $\varphi^{-1}(0,1)$, $\varphi^{-1}(1,0)$, and $\varphi^{-1}(1,1)$. Now we easily see that \[\varphi^{-1}(0,0)=0\] and \[\varphi^{-1}(1,1)=1.\] Noting that $625\equiv 1\pmod{16}$, it follows that \[\varphi^{-1}(1,0)=625.\] To compute $\varphi^{-1}(0,1)$, note that \[(0,1)=15(1,0)+(1,1)\] in \[\mathbb \mathbb Z/16\mathbb Z\times\mathbb Z/625\mathbb Z,\] so since $\varphi^{-1}$ is linear in its arguments (by virtue of being an isomorphism), \[\varphi^{-1}(0,1)=15\varphi^{-1}(1,0)+\varphi^{-1}(1,1)=15\times 625+1=9376.\]

The four candidate digit strings $abcd$ are then $0000,0001,0625,9376$. Of those, only $9376$ has nonzero first digit, and therefore the answer is $\boxed{937}$.

Solution 4 (semi-bashing)

  • Note - $\overline{abcd}$ means the number formed when the digits represented by $a$, $b$, $c$, and $d$ are substituted in. $\overline{abcd}\ne a\times b\times c\times d$.

WLOG, we can assume that $N$ is a 4-digit integer $\overline{abcd}$. Note that the only $d$ that will satisfy $N$ will also satisfy $d^2\equiv d\pmod{10}$, as the units digit of $\overline{abcd}^2$ is affected only by $d$, regardless of $a$, $b$, or $c$.

By checking the numbers 0-9, we see that the only possible values of $d$ are $d=0, 1, 5, 6$.

Now, we seek to find $c$. Note that the only $\overline{cd}$ that will satisfy $N$ will also satisfy $\overline{cd}^2 \equiv \overline{cd}\pmod{100}$, by the same reasoning as above - the last two digits of $\overline{abcd}^2$ are only affected by $c$ and $d$. As we already have narrowed choices for $d$, we start reasoning out.

First, we note that if $d=0$, then $c=0$, as a number ending in 0, and therefore divisible by 10, is squared, the result is divisible by 100, meaning it ends in two 0's. However, if $N$ ends in $00$, then recursively, $a$ and $b$ must be $0$, as a number divisible by 100 squared ends in four zeros. As $a$ cannot be 0, we throw out this possibility, as the only solution in this case is $0$.

Now, let's assume that $d=1$. $\overline{cd}$ is equal to $10c + d = 10c + 1$. Squaring this gives $100c^2 + 20c + 1$, and when modulo 100 is taken, it must equal $10c + 1$. As $c$ is an integer, $100c^2$ must be divisible by 100, so $100c^2+20c+1 \equiv 20c + 1\pmod{100}$, which must be equivalent to $10c + 1$. Note that this is really $\overline{(2c)1}$ and $\overline{c1}$, and comparing the 10's digits. So really, we're just looking for when the units digit of $2c$ and $c$ are equal, and a quick check reveals that this is only true when $c=0$.However, if we extend this process to find $b$ and $a$, we'd find that they are also 0. The only solution in this case is $1$, and since $a=0$ here, this is not our solution. Therefore, there are no valid solutions in this case.

Let's assume that $d=5$. Note that $(10c + 5)^2 = 100c^2 + 100c + 25$, and when modulo $100$ is taken, $25$ is the remainder. So all cases here have squares that end in 25, so $\overline{cd}=25$ is our only case here. A quick check reveals that $25^2=625$, which works for now.

Now, let $d=6$. Note that $(10c + 6)^2 = 100c^2 + 120c + 36$. Taking modulo 100, this reduces to $20c+36$, which must be equivalent to $10c+6$. Again, this is similar to $\overline{(2c+3)6}$ and $\overline{c6}$, so we see when the units digits of $2c+3$ and $c$ are equal. To make checking faster, note that $2c$ is necessarily even, so $2c+3$ is necessarily odd, so $c$ must be odd. Checking all the odds reveals that only $c=3$ works, so this case gives $76$. Checking quickly $76^2 = 5776$, which works for now.

Now, we find $b$, given two possibilities for $\overline{cd}$.

Start with $\overline{cd} = 25$. $\overline{bcd} = 100b + \overline{cd} = 100b + 25.$ Note that if we square this, we get $10000b^2 + 5000b + 625$, which should be equivalent to $100b + 25$ modulo 1000. Note that, since $b$ is an integer, $10000b^2 + 5000 + 625$ simplifies modulo 1000 to $625$. Therefore, the only $\overline{bcd}$ that works here is $625$. $625^2 = 390625$.

Now, assume that $\overline{cd}=76$. We have $100b + 76$, and when squared, becomes $10000b^2 + 15200b + 5776$, which, modulo 1000, should be equivalent to $100b+76$. Reducing $10000b^2 + 15200b + 5776$ modulo $1000$ gives $200b + 776$. Using the same technique as before, we must equate the hundreds digit of $\overline{(2b+7)76}$ to $\overline{b76}$, or equate the units digit of $2b+7$ and $b$. Since $2b+7$ is necessarily odd, any possible $b$'s must be odd. A quick check reveals that $b=3$ is the only solution, so we get a solution of $376$. $376^2 = 141376$.

Finally, we solve for $a$. Start with $\overline{bcd}=625$. We have $1000a + 625$, which, squared, gives \[1000000a^2 + 1250000a + 390625,\] and reducing modulo 10000 gives simply 625. So $\overline{abcd}=625$. However, that makes $a=0$. Therefore, no solutions exist in this case.

We turn to our last case, $\overline{bcd}=376$. We have \[1000a + 376^2 = 1000000a^2 + 752000a + 141376,\] and reducing modulo $10000$ gives $2000a + 1376$, which must be equivalent to $1000a + 376$. So we must have $\overline{(2a+1)376}$ being equivalent to $\overline{a376}$ modulo 1000. So, the units digit of $2a+1$ must be equal to $a$. Since $2a+1$ is odd, $a$ must be odd. Lo and behold, the only possibility for $a$ is $a=3$. Therefore, $\overline{abcd}=9376$, so our answer is $\boxed{937}$.

See also

2014 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 7
Followed by
Problem 9
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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