Difference between revisions of "2017 AIME I Problems/Problem 2"
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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>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 | ||
+ | |||
+ | ==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 03:08, 5 January 2024
Contents
Problem 2
When each of ,
, and
is divided by the positive integer
, the remainder is always the positive integer
. When each of
,
, and
is divided by the positive integer
, the remainder is always the positive integer
. Find
.
Solution
Let's work on both parts of the problem separately. First, We take the difference of
and
, and also of
and
. We find that they are
and
, respectively. Since the greatest common divisor of the two differences is
(and the only one besides one), it's safe to assume that
.
Then, we divide by
, and it's easy to see that
. Dividing
and
by
also yields remainders of
, which means our work up to here is correct.
Doing the same thing with ,
, and
, the differences between
and
and
are
and
, respectively. Since the only common divisor (besides
, of course) is
,
. Dividing all
numbers by
yields a remainder of
for each, so
. Thus,
.
Solution 2
We know that and
where
are integers.
Subtracting the first two, the first and third, and the last two, we get and
We know that and
must be integers, so all the numbers are divisible by
Factorizing the numbers, we get and
We see that all these have a factor of 17, so
Finding the remainder when is divided by
we get
Doing the same thing for the next three numbers, we get
~solasky
Solution 3 (Sol. 1 but possibly more clear)
As in Solution 1, we are given and
Tackling the first equation, we can simply look at
. We subtract
from each component of the congruency to get
. Thus, we know that
and
must both be divisible by
. The only possible
, in this case, become
and
; obviously,
, so we know
. We go back to the original equation, plug in
, and we find that
.
Similarly, we can subtract out the smallest value in the second congruency, . We end up with
. Again, we find that
or
, so
. We also find that
.
Thus, our answer is .
~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.