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

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(Solution)
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==Solution==
 
==Solution==
Let's tackle the first part of the problem first. We can safely assume: <cmath>702 = xm + r</cmath> <cmath>787 = ym + r</cmath> <cmath>855 = zm + r</cmath>
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Let's work on both parts of the problem separately. First, <cmath>855 \equiv 787 \equiv 702 \equiv r \pmod{m}.</cmath> We take the difference of <math>855</math> and <math>787</math>, and also of <math>787</math> and <math>702</math>. We find that they are <math>85</math> and <math>68</math>, respectively. Since the greatest common divisor of the two differences is <math>17</math> (and the only one besides one), it's safe to assume that <math>m = 17</math>.
Now, if we subtract two values: <cmath>787-702 = 85 = 17\cdot5</cmath>
 
which also equals <cmath>(ym+r)-(xm+r) = m\cdot(y-x)</cmath>
 
Similarly, <cmath>855-787 = 68 = 17\cdot4; (zm+r)-(ym+r) = m\cdot(z-y)</cmath>
 
Since <math>17</math> is the only common factor, we can assume that <math>m=17</math>, and through simple division, that <math>r=5</math>.
 
  
Using the same method on the second half: <cmath>412 = an + s</cmath> <cmath>722 = bn + s</cmath> <cmath>815 = cn + s</cmath>
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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. <cmath>722-412 = 310 = 31\cdot10; (bn+s)-(an+s) = n\cdot(b-a)</cmath> <cmath>815-722 = 93 = 31\cdot3; (cn+s)-(bn+s) = n\cdot(c-b)</cmath>
 
The common factor is <math>31</math>, so <math>n=31</math> and through division, <math>s=9</math>.
 
 
 
The answer is <math>m+n+r+s = 17+31+5+9 = \boxed{62}</math>
 
  
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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{62}</math>.
  
 
==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}}

Revision as of 23:32, 7 August 2017

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 $722$ 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, $m + n + r + s = 17 + 5 + 31 + 9 = \boxed{62}$.

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

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