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Difference between revisions of "2012 AIME I Problems/Problem 10"

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==Problem 10==
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==Problem==
  
 
Let <math>\mathcal{S}</math> be the set of all perfect squares whose rightmost three digits in base <math>10</math> are <math>256</math>. Let <math>\mathcal{T}</math> be the set of all numbers of the form <math>\frac{x-256}{1000}</math>, where <math>x</math> is in <math>\mathcal{S}</math>. In other words, <math>\mathcal{T}</math> is the set of numbers that result when the last three digits of each number in <math>\mathcal{S}</math> are truncated. Find the remainder when the tenth smallest element of <math>\mathcal{T}</math> is divided by <math>1000</math>.
 
Let <math>\mathcal{S}</math> be the set of all perfect squares whose rightmost three digits in base <math>10</math> are <math>256</math>. Let <math>\mathcal{T}</math> be the set of all numbers of the form <math>\frac{x-256}{1000}</math>, where <math>x</math> is in <math>\mathcal{S}</math>. In other words, <math>\mathcal{T}</math> is the set of numbers that result when the last three digits of each number in <math>\mathcal{S}</math> are truncated. Find the remainder when the tenth smallest element of <math>\mathcal{T}</math> is divided by <math>1000</math>.
  
 
== Solution 1==
 
== Solution 1==
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It is apparent that for a perfect square <math>s^2</math> to satisfy the constraints, we must have <math>s^2 - 256 = 1000n</math> or <math>(s+16)(s-16) = 1000n.</math> Now in order for <math>(s+16)(s-16)</math> to be a multiple of <math>1000,</math> at least one of <math>s+16</math> and <math>s-16</math> must be a multiple of <math>5,</math> and since <math>s+16</math> and <math>s-16</math> are in different residue classes mod <math>5,</math> one term must have all the factors of <math>5</math> and thus must be a multiple of <math>125.</math> Furthermore, each of <math>s+16</math> and <math>s-16</math> must have at least two factors of <math>2,</math> since otherwise <math>(s+16)(s-16)</math> could not possibly be divisible by <math>8.</math> So therefore the conditions are satisfied if either <math>s+16</math> or <math>s-16</math> is divisible by <math>500,</math> or equivalently if <math>s = 500n \pm 16.</math> Counting up from <math>n=0</math> to <math>n=5,</math> we see that the tenth value of <math>s</math> is <math>500 \cdot 5 - 16 = 2484</math> and thus the corresponding element in <math>\mathcal{T}</math> is <math>\frac{2484^2 - 256}{1000} = 6170 \rightarrow \boxed{170.}</math>
+
It is apparent that for a perfect square <math>s^2</math> to satisfy the constraints, we must have <math>s^2 - 256 = 1000n</math> or <math>(s+16)(s-16) = 1000n.</math> Now in order for <math>(s+16)(s-16)</math> to be a multiple of <math>1000,</math> at least one of <math>s+16</math> and <math>s-16</math> must be a multiple of <math>5,</math> and since <math>s+16 \not\equiv s-16 \pmod{5},</math> one term must have all the factors of <math>5</math> and thus must be a multiple of <math>125.</math> Furthermore, each of <math>s+16</math> and <math>s-16</math> must have at least two factors of <math>2,</math> since otherwise <math>(s+16)(s-16)</math> could not possibly be divisible by <math>8.</math> So therefore the conditions are satisfied if either <math>s+16</math> or <math>s-16</math> is divisible by <math>500,</math> or equivalently if <math>s = 500n \pm 16.</math> Counting up from <math>n=0</math> to <math>n=5,</math> we see that the tenth value of <math>s</math> is <math>500 \cdot 5 - 16 = 2484</math> and thus the corresponding element in <math>\mathcal{T}</math> is <math>\frac{2484^2 - 256}{1000} = 6170 \rightarrow \boxed{170.}</math>
  
 
== Solution 2==
 
== Solution 2==
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Notice that is <math>16^2=256</math>, <math>1016^2</math> ends in <math>256</math>. In general, if <math>x^2</math> ends in <math>256</math>, <math>(x+1000)^2=x^2+2000x+1000000</math> ends in 256 because <math>2000x >1000</math> and <math>2000000 > 1000</math>. It is clear that we want all numbers whose squares end in <math>256</math> that are less than <math>1000</math>.
+
Notice that is <math>16^2=256</math>, <math>1016^2</math> ends in <math>256</math>. In general, if <math>x^2</math> ends in <math>256</math>, <math>(x+1000)^2=x^2+2000x+1000000</math> ends in 256 because <math>1000|2000x</math> and <math>1000|2000000</math>. It is clear that we want all numbers whose squares end in <math>256</math> that are less than <math>1000</math>.
  
 
Firstly, we know the number has to end in a <math>4</math> or a <math>6</math>. Let's look at the ones ending in <math>6</math>.
 
Firstly, we know the number has to end in a <math>4</math> or a <math>6</math>. Let's look at the ones ending in <math>6</math>.
  
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Assume that the second digit of the three digit number is <math>0</math>. We find that the last <math>3</math> digits of <math>\overline{a06}^2</math> is in the form <math>12a \cdot 100 + 3 \cdot 10 + 6</math>. However, the last two digits need to be a <math>56</math>. Thus, similarly trying all numbers from <math>0</math> to <math>10</math>, we see that only 1 for the middle digit works. Carrying out the multiplication, we see that the last <math>3</math> digits of <math>\overline{a06}^2</math> is in the form <math>(12a + 2) \cdot 100 + 5 \cdot 10 + 6</math>. We want <math>(12a + 2)\pmod{10}</math> to be equal to <math>2</math>. Thus, we see that a is <math>0</math> or <math>5</math>. Thus, the numbers that work in this case are <math>016</math> and <math>516</math>.  
+
Assume that the second digit of the three digit number is <math>0</math>. We find that the last <math>3</math> digits of <math>\overline{a06}^2</math> is in the form <math>12a \cdot 100 + 3 \cdot 10 + 6</math>. However, the last two digits need to be a <math>56</math>. Thus, similarly trying all numbers from <math>0</math> to <math>10</math>, we see that only 1 for the middle digit works. Carrying out the multiplication, we see that the last <math>3</math> digits of <math>\overline{a16}^2</math> is in the form <math>(12a + 2) \cdot 100 + 5 \cdot 10 + 6</math>. We want <math>(12a + 2)\pmod{10}</math> to be equal to <math>2</math>. Thus, we see that a is <math>0</math> or <math>5</math>. Thus, the numbers that work in this case are <math>016</math> and <math>516</math>.  
  
 
Next, let's look at the ones ending in <math>4</math>. Carrying out a similar technique as above, we see that the last <math>3</math> digits of <math>\overline{a84}^2</math> is in the form <math>((8a+10) \cdot 100+ 5 \cdot 10 + 6</math>. We want <math>(8a + 10)\pmod{10}</math> to be equal to <math>2</math>. We see that only <math>4</math> and <math>9</math> work. Thus, we see that only <math>484</math> and <math>984</math> work.
 
Next, let's look at the ones ending in <math>4</math>. Carrying out a similar technique as above, we see that the last <math>3</math> digits of <math>\overline{a84}^2</math> is in the form <math>((8a+10) \cdot 100+ 5 \cdot 10 + 6</math>. We want <math>(8a + 10)\pmod{10}</math> to be equal to <math>2</math>. We see that only <math>4</math> and <math>9</math> work. Thus, we see that only <math>484</math> and <math>984</math> work.
  
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We order these numbers to get  <math>16</math>, <math>484</math>, <math>516</math> <math>984</math>. We want the <math>10th</math> number in order which is <math>2484^2 = 6170256</math>. <math>\boxed{170}</math>
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We order these numbers to get  <math>16</math>, <math>484</math>, <math>516</math>, <math>984</math>... We want the <math>10th</math> number in order which is <math>2484^2 = 6\boxed{170}256</math>.
 +
 
 +
== Solution 3==
 +
The condition implies <math>x^2\equiv 256 \pmod{1000}</math>. Rearranging and factoring, <cmath>(x-16)(x+16)\equiv 0\pmod {1000}.</cmath>
 +
This can be expressed with the system of congruences
 +
<cmath>
 +
\begin{cases}
 +
(x-16)(x+16)\equiv 0\pmod{125} \\ (x-16)(x+16)\equiv 0\pmod{8}
 +
\end{cases}
 +
</cmath>
 +
Observe that <math>x\equiv {109} \pmod {125}</math> or <math>x\equiv{16}\pmod {125}</math>. Similarly, it can be seen that <math>x\equiv{0}\pmod{8}</math> or <math>x\equiv{4}\pmod{8}</math>. By CRT, there are four solutions to this modulo <math>1000</math> (one for each case e.g. <math>x\equiv{109}</math> and <math>x\equiv{0}</math> or <math>x\equiv{125}</math> and <math>x\equiv{4}</math>. These solutions are (working modulo <math>1000</math>)
 +
<cmath>
 +
\begin{cases}
 +
x=16 \\ x=484 \\ x= 516 \\ x=984
 +
\end{cases}
 +
</cmath>
 +
The tenth solution is <math>x=2484,</math> which gives an answer of <math>\boxed{170}</math>.
 +
 
 +
== Solution 4 ==
 +
An element in S can be represented by <math>y^2 = 1000a + 256</math>, where <math>y^2</math> is the element in S. Since the right hand side must be even, we let <math>y = 2y_1</math> and substitute to get <math>y_1^2 = 250a + 64</math>. However, the right hand side is still even, so we make the substitution <math>y_1 = 2y_2</math> to get <math>y_2^2 = 125a/2 + 16</math>. Because both sides must be an integer, we know that <math>a = 2a_1</math> for some integer <math>a_1</math>. Our equation then becomes <math>y_2^2 = 125a_1 + 16</math>, and we can simplify no further.
 +
 
 +
Rearranging terms, we get <math>y_2^2 - 16 = 125a_1</math>, whence difference of squares gives <math>(y_2 + 4)(y_2 - 4) = 125a_1</math>. Note that this equation tells us that one of <math>y_2 + 4</math> and <math>y_2 - 4</math> contains a nonnegative multiple of <math>125</math>. Hence, listing out the smallest possible values of <math>y_2</math>, we have <math>y_2 = 4, 121, 129, \cdots, 621</math>. The tenth term is <math>621</math>, whence <math>y = 4y_2 = 2484</math>. The desired result can then be calculated to be <math>\boxed{170}</math>. - Spacesam
 +
 
 +
== Solution 5 (Similar to Solution 4) ==
 +
From the conditions, we can let every element in <math>\mathcal{S}</math> be written as <math>y^2=1000x+256</math>, where <math>x</math> and <math>y</math> are integers. Since there are no restrictions on <math>y</math>, we let <math>y_1</math> be equal to <math>y+16</math> (<math>y-16</math> works as well). Then the <math>256</math> cancels out and we're left with <cmath>y_1^2+32y_1=1000x</cmath> which can be factored as <cmath>y_1(y_1+32)=1000x</cmath> Since the RHS is even, <math>y_1</math> must be even, so we let <math>y_1=2y_2</math>, to get <cmath>y_2(y_2+16)=250x</cmath> Again, because the RHS is even, the LHS must be even too, so substituting <math>y_3=\frac{1}{2}y_2</math> we have <cmath>y_3(y_3+8)=125\cdot\frac{1}{2}x</cmath> Since the LHS is an integer, the RHS must thus be an integer, so substituting <math>x=2x_1</math> we get <cmath>y_3(y_3+8)=125x_1</cmath> Then we can do casework on the values of <math>y_3</math>, as only one of <math>y_3</math> and <math>y_3+8</math> can be multiples of <math>125</math>
 +
 
 +
'''Case 1''': <math>125|y_3</math>
 +
 
 +
Since we're trying to find the values of <math>x_1</math>, we can let <math>y_4=\frac{1}{125}y_3</math>, to get <cmath>x_1=y_4(125y_4+8)</cmath> or <cmath>x=2y_4(125y_4+8)</cmath>
 +
 
 +
'''Case 2''': <math>125|y_3+8</math>
 +
 
 +
Similar to Case 1, only the equation is <cmath>x=2y_4(125y_4-8)</cmath>
 +
 
 +
In whole, the values of <math>x</math> (i.e. the elements in <math>\mathcal{T}</math>) are of the form <cmath>x=2k(125k\pm8)</cmath> where <math>k</math> is any integer. It can easily be seen that if <math>k<0</math>, then <math>x</math> is negative, thus <math>k\geq0</math>. Also, note that when <math>k=0</math>, there is only one value, because one of the factors is <math>0</math> (<math>k</math>). Thus the <math>10^{th}</math> smallest number in the set <math>\mathcal{T}</math> is when the <math>\pm</math> sign takes the minus side and <math>k=5</math>, giving <math>6170</math>, so the answer is <math>\boxed{170}</math>
 +
 
 +
== Video Solution by Richard Rusczyk ==
 +
 
 +
https://artofproblemsolving.com/videos/amc/2012aimei/349
 +
 
 +
~ dolphin7
 +
 
 +
==Video Solution==
 +
 
 +
https://www.youtube.com/watch?v=caCELHibbIE&list=PLOaAlyCEsUTbA2v1gRyEAB_gxk7NiWG3v&index=6&t=0s
 +
 
 +
~Shreyas S
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=2012|n=I|num-b=9|num-a=11}}
 
{{AIME box|year=2012|n=I|num-b=9|num-a=11}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 17:08, 7 September 2024

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 $\overline{a16}^2$ 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 $\boxed{170}$

Video Solution by Richard Rusczyk

https://artofproblemsolving.com/videos/amc/2012aimei/349

~ dolphin7

Video Solution

https://www.youtube.com/watch?v=caCELHibbIE&list=PLOaAlyCEsUTbA2v1gRyEAB_gxk7NiWG3v&index=6&t=0s

~Shreyas S

See also

2012 AIME I (Problems • Answer Key • Resources)
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