# 2012 AIME I Problems/Problem 10

(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

## Problem

Let $\mathcal{S}$ be the set of all perfect squares whose rightmost three digits in base $10$ are $256$. Let $\mathcal{T}$ be the set of all numbers of the form $\frac{x-256}{1000}$, where $x$ is in $\mathcal{S}$. In other words, $\mathcal{T}$ is the set of numbers that result when the last three digits of each number in $\mathcal{S}$ are truncated. Find the remainder when the tenth smallest element of $\mathcal{T}$ is divided by $1000$.

## Solution 1

It is apparent that for a perfect square $s^2$ to satisfy the constraints, we must have $s^2 - 256 = 1000n$ or $(s+16)(s-16) = 1000n.$ Now in order for $(s+16)(s-16)$ to be a multiple of $1000,$ at least one of $s+16$ and $s-16$ must be a multiple of $5,$ and since $s+16 \not\equiv s-16 \pmod{5},$ one term must have all the factors of $5$ and thus must be a multiple of $125.$ Furthermore, each of $s+16$ and $s-16$ must have at least two factors of $2,$ since otherwise $(s+16)(s-16)$ could not possibly be divisible by $8.$ So therefore the conditions are satisfied if either $s+16$ or $s-16$ is divisible by $500,$ or equivalently if $s = 500n \pm 16.$ Counting up from $n=0$ to $n=5,$ we see that the tenth value of $s$ is $500 \cdot 5 - 16 = 2484$ and thus the corresponding element in $\mathcal{T}$ is $\frac{2484^2 - 256}{1000} = 6170 \rightarrow \boxed{170.}$

## Solution 2

Notice that is $16^2=256$, $1016^2$ ends in $256$. In general, if $x^2$ ends in $256$, $(x+1000)^2=x^2+2000x+1000000$ ends in 256 because $1000|2000x$ and $1000|2000000$. It is clear that we want all numbers whose squares end in $256$ that are less than $1000$.

Firstly, we know the number has to end in a $4$ or a $6$. Let's look at the ones ending in $6$.

Assume that the second digit of the three digit number is $0$. We find that the last $3$ digits of $\overline{a06}^2$ is in the form $12a \cdot 100 + 3 \cdot 10 + 6$. However, the last two digits need to be a $56$. Thus, similarly trying all numbers from $0$ to $10$, we see that only 1 for the middle digit works. Carrying out the multiplication, we see that the last $3$ digits of $\overline{a06}^2$ is in the form $(12a + 2) \cdot 100 + 5 \cdot 10 + 6$. We want $(12a + 2)\pmod{10}$ to be equal to $2$. Thus, we see that a is $0$ or $5$. Thus, the numbers that work in this case are $016$ and $516$.

Next, let's look at the ones ending in $4$. Carrying out a similar technique as above, we see that the last $3$ digits of $\overline{a84}^2$ is in the form $((8a+10) \cdot 100+ 5 \cdot 10 + 6$. We want $(8a + 10)\pmod{10}$ to be equal to $2$. We see that only $4$ and $9$ work. Thus, we see that only $484$ and $984$ work.

We order these numbers to get $16$, $484$, $516$, $984$... We want the $10th$ number in order which is $2484^2 = 6\boxed{170}256$.

## Solution 3

The condition implies $x^2\equiv 256 \pmod{1000}$. Rearranging and factoring, $$(x-16)(x+16)\equiv 0\pmod {1000}.$$ This can be expressed with the system of congruences $$\begin{cases} (x-16)(x+16)\equiv 0\pmod{125} \\ (x-16)(x+16)\equiv 0\pmod{8} \end{cases}$$ Observe that $x\equiv {109} \pmod {125}$ or $x\equiv{16}\pmod {125}$. Similarly, it can be seen that $x\equiv{0}\pmod{8}$ or $x\equiv{4}\pmod{8}$. By CRT, there are four solutions to this modulo $1000$ (one for each case e.g. $x\equiv{109}$ and $x\equiv{0}$ or $x\equiv{125}$ and $x\equiv{4}$. These solutions are (working modulo $1000$) $$\begin{cases} x=16 \\ x=484 \\ x= 516 \\ x=984 \end{cases}$$ The tenth solution is $x=2484,$ which gives an answer of $\boxed{170}$.

## Solution 4

An element in S can be represented by $y^2 = 1000a + 256$, where $y^2$ is the element in S. Since the right hand side must be even, we let $y = 2y_1$ and substitute to get $y_1^2 = 250a + 64$. However, the right hand side is still even, so we make the substitution $y_1 = 2y_2$ to get $y_2^2 = 125a/2 + 16$. Because both sides must be an integer, we know that $a = 2a_1$ for some integer $a_1$. Our equation then becomes $y_2^2 = 125a_1 + 16$, and we can simplify no further.

Rearranging terms, we get $y_2^2 - 16 = 125a_1$, whence difference of squares gives $(y_2 + 4)(y_2 - 4) = 125a_1$. Note that this equation tells us that one of $y_2 + 4$ and $y_2 - 4$ contains a nonnegative multiple of $125$. Hence, listing out the smallest possible values of $y_2$, we have $y_2 = 4, 121, 129, \cdots, 621$. The tenth term is $621$, whence $y = 4y_2 = 2484$. The desired result can then be calculated to be $\boxed{170}$. - Spacesam

## Solution 5 (Similar to Solution 4)

From the conditions, we can let every element in $\mathcal{S}$ be written as $y^2=1000x+256$, where $x$ and $y$ are integers. Since there are no restrictions on $y$, we let $y_1$ be equal to $y+16$ ($y-16$ works as well). Then the $256$ cancels out and we're left with $$y_1^2+32y_1=1000x$$ which can be factored as $$y_1(y_1+32)=1000x$$ Since the RHS is even, $y_1$ must be even, so we let $y_1=2y_2$, to get $$y_2(y_2+16)=250x$$ Again, because the RHS is even, the LHS must be even too, so substituting $y_3=\frac{1}{2}y_2$ we have $$y_3(y_3+8)=125\cdot\frac{1}{2}x$$ Since the LHS is an integer, the RHS must thus be an integer, so substituting $x=2x_1$ we get $$y_3(y_3+8)=125x_1$$ Then we can do casework on the values of $y_3$, as only one of $y_3$ and $y_3+8$ can be multiples of $125$

Case 1: $125|y_3$

Since we're trying to find the values of $x_1$, we can let $y_4=\frac{1}{125}y_3$, to get $$x_1=y_4(125y_4+8)$$ or $$x=2y_4(125y_4+8)$$

Case 2: $125|y_3+8$

Similar to Case 1, only the equation is $$x=2y_4(125y_4-8)$$

In whole, the values of $x$ (i.e. the elements in $\mathcal{T}$) are of the form $$x=2k(125k\pm8)$$ where $k$ is any integer. It can easily be seen that if $k<0$, then $x$ is negative, thus $k\geq0$. Also, note that when $k=0$, there is only one value, because one of the factors is $0$ ($k$). Thus the $10^{th}$ smallest number in the set $\mathcal{T}$ is when the $\pm$ sign takes the minus side and $k=5$, giving $6170$, so the answer is $\boxed{170}$

~ dolphin7

~Shreyas S