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2012 AIME I Problems/Problem 10

Revision as of 13:57, 6 July 2015 by Daniellionyang (talk | contribs) (→‎Solution)

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 $2484^2 = 6170256$. $\boxed{170}$

See also

2012 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 9 		Followed by
Problem 11
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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