Difference between revisions of "2017 AIME I Problems/Problem 2"

Latest revision as of 02:08, 5 January 2024

Problem 2

When each of $702$ , $787$ , and $855$ is divided by the positive integer $m$ , the remainder is always the positive integer $r$ . When each of $412$ , $722$ , and $815$ is divided by the positive integer $n$ , the remainder is always the positive integer $s \neq r$ . Find $m+n+r+s$ .

Solution

Let's work on both parts of the problem separately. First, $855 \equiv 787 \equiv 702 \equiv r \pmod{m}.$ We take the difference of $855$ and $787$ , and also of $787$ and $702$ . We find that they are $85$ and $68$ , respectively. Since the greatest common divisor of the two differences is $17$ (and the only one besides one), it's safe to assume that $m = 17$ .

Then, we divide $855$ by $17$ , and it's easy to see that $r = 5$ . Dividing $787$ and $702$ by $17$ also yields remainders of $5$ , which means our work up to here is correct.

Doing the same thing with $815$ , $722$ , and $412$ , the differences between $815$ and $722$ and $412$ are $310$ and $93$ , respectively. Since the only common divisor (besides $1$ , of course) is $31$ , $n = 31$ . Dividing all $3$ numbers by $31$ yields a remainder of $9$ for each, so $s = 9$ . Thus, $m + n + r + s = 17 + 5 + 31 + 9 = \boxed{062}$ .

Solution 2

We know that $702 = am + r, 787 = bm + r,$ and $855 = cm+r$ where $a-c$ are integers.

Subtracting the first two, the first and third, and the last two, we get $85 = (b-a)m, 153=(c-a)m,$ and $68=(c-b)m.$

We know that $b-a, c-a$ and $c-b$ must be integers, so all the numbers are divisible by $m.$

Factorizing the numbers, we get $85 = 5 \cdot 17, 153 = 3^2 \cdot 17,$ and $68 = 2^2 \cdot 17.$ We see that all these have a factor of 17, so $m=17.$

Finding the remainder when $702$ is divided by $17,$ we get $n=5.$

Doing the same thing for the next three numbers, we get $17 + 5 + 31 + 9 = \boxed{062}$

~solasky

Solution 3 (Sol. 1 but possibly more clear)

As in Solution 1, we are given $855\equiv787\equiv702\equiv r\pmod{m}$ and $815\equiv722\equiv412\equiv s\pmod{n}.$ Tackling the first equation, we can simply look at $855\equiv787\equiv702\pmod{m}$ . We subtract $702$ from each component of the congruency to get $153\equiv85\equiv0\pmod{m}$ . Thus, we know that $153$ and $85$ must both be divisible by $m$ . The only possible $m$ , in this case, become $17$ and $1$ ; obviously, $m\neq1$ , so we know $m=17$ . We go back to the original equation, plug in $m$ , and we find that $r=5$ .

Similarly, we can subtract out the smallest value in the second congruency, $412$ . We end up with $403\equiv310\equiv0\pmod{n}$ . Again, we find that $n=31$ or $1$ , so $n=31$ . We also find that $s=9$ .

Thus, our answer is $\boxed{062}$ .

~Technodoggo

Video Solution

https://youtu.be/BiiKzctXDJg ~Shreyas S

@@ Line 23: / Line 23: @@
 ~solasky
+==Solution 3 (Sol. 1 but possibly more clear)==
+As in Solution 1, we are given <math>855\equiv787\equiv702\equiv r\pmod{m}</math> and <math>815\equiv722\equiv412\equiv s\pmod{n}.</math> Tackling the first equation, we can simply look at <math>855\equiv787\equiv702\pmod{m}</math>. We subtract <math>702</math> from each component of the congruency to get <math>153\equiv85\equiv0\pmod{m}</math>. Thus, we know that <math>153</math> and <math>85</math> must both be divisible by <math>m</math>. The only possible <math>m</math>, in this case, become <math>17</math> and <math>1</math>; obviously, <math>m\neq1</math>, so we know <math>m=17</math>. We go back to the original equation, plug in <math>m</math>, and we find that <math>r=5</math>.
+Similarly, we can subtract out the smallest value in the second congruency, <math>412</math>. We end up with <math>403\equiv310\equiv0\pmod{n}</math>. Again, we find that <math>n=31</math> or <math>1</math>, so <math>n=31</math>. We also find that <math>s=9</math>.
+Thus, our answer is <math>\boxed{062}</math>.
+~Technodoggo
 ==Video Solution==

2017 AIME I (Problems • Answer Key • Resources)
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