Difference between revisions of "2017 AIME I Problems/Problem 2"
m (replace ans "62" with "062") |
|||
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>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{ | + | 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>. |
==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 22:04, 3 March 2020
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, .
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.