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

m (Solution)
m (Solution 2)
 
(7 intermediate revisions by 5 users not shown)
Line 7: Line 7:
 
Then, we divide <math>855</math> by <math>17</math>, and it's easy to see that <math>r = 5</math>. Dividing <math>787</math> and <math>702</math> by <math>17</math> also yields remainders of <math>5</math>, which means our work up to here is correct.
 
Then, we divide <math>855</math> by <math>17</math>, and it's easy to see that <math>r = 5</math>. Dividing <math>787</math> and <math>702</math> by <math>17</math> also yields remainders of <math>5</math>, which means our work up to here is correct.
  
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>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>.
+
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{062}</math>
 +
 
 +
~solasky
 +
 
 +
==Video Solution==
 +
 
 +
https://youtu.be/BiiKzctXDJg ~Shreyas S
  
 
==See Also==
 
==See Also==
 
{{AIME box|year=2017|n=I|num-b=1|num-a=3}}
 
{{AIME box|year=2017|n=I|num-b=1|num-a=3}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 16:22, 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 $17 + 5 + 31 + 9 = \boxed{062}$

~solasky

Video Solution

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

See Also

2017 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 3
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS