Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m
(added sol 3)
 
(13 intermediate revisions by 9 users not shown)
Line 3: Line 3:
  
 
==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>
+
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>
+
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>
+
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>.
  
−
~IYN~
+
==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
 +
 
 +
==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==
 +
 
 +
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 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

See Also

2017 AIME I (Problems • Answer Key • Resources)
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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2017_AIME_I_Problems/Problem_2&oldid=209950"