2012 AIME I Problems/Problem 10

Problem 10

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 $s+16$ and $s-16$ are in different residue classes mod $5,$ 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 $2000x >1000$ and $2000000 > 1000$ . 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 $2484^2 = 6170256$ . $\boxed{170}$

Solution 2

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 $\boxed{170}$ .

2012 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 9	Followed by Problem 11
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2012 AIME I Problems/Problem 10

Contents

Problem 10

Solution 1

Solution 2

Solution 2

See also