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

Revision as of 16:21, 9 March 2021

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 $17 + 5 + 31 + 9 = \boxed{62}$

Video Solution

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

@@ Line 8: / Line 8: @@
 Doing the same thing with <math>815</math>, <math>722</math>, and <math>412</math>, the differences between <math>815</math> and <math>722</math> and <math>412</math> are <math>310</math> and <math>93</math>, respectively. Since the only common divisor (besides <math>1</math>, of course) is <math>31</math>, <math>n = 31</math>. Dividing all <math>3</math> numbers by <math>31</math> yields a remainder of <math>9</math> for each, so <math>s = 9</math>. Thus, <math>m + n + r + s = 17 + 5 + 31 + 9 = \boxed{062}</math>.
+==Solution 2==
+We know that <math>702 = am + r, 787 = bm + r,</math> and <math>855 = cm+r</math> where <math>a-c</math> are integers.
+Subtracting the first two, the first and third, and the last two, we get <math>85 = (b-a)m, 153=(c-a)m,</math> and <math>68=(c-b)m.</math>
+We know that <math>b-a, c-a</math> and <math>c-b</math> must be integers, so all the numbers are divisible by <math>m.</math>
+Factorizing the numbers, we get <math>85 = 5 \cdot 17, 153 = 3^2 \cdot 17,</math> and <math>68 = 2^2 \cdot 17.</math> We see that all these have a factor of 17, so <math>m=17.</math>
+Finding the remainder when <math>702</math> is divided by <math>17,</math> we get <math>n=5.</math>
+Doing the same thing for the next three numbers, we get <math>17 + 5 + 31 + 9 = \boxed{62}</math>
 ==Video Solution==

2017 AIME I (Problems • Answer Key • Resources)
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