Difference between revisions of "1999 AHSME Problems/Problem 25"
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== Solution 3 (The Easiest and Most Intuitive) == | == Solution 3 (The Easiest and Most Intuitive) == | ||
− | Let's clear the fractions: <cmath>\frac{5}{7}=\frac{2520a_2+840a_3+210a_4+42a_5+7a_6+a_7}{7!}</cmath> | + | Let's clear up the fractions: <cmath>\frac{5}{7}=\frac{2520a_2+840a_3+210a_4+42a_5+7a_6+a_7}{7!}</cmath> |
<cmath>3600=2520a_2+840a_3+210a_4+42a_5+7a_6+a_7</cmath> | <cmath>3600=2520a_2+840a_3+210a_4+42a_5+7a_6+a_7</cmath> | ||
Notice that if we divide everything by <math>7</math> then we would have:<cmath>514+\frac{2}{7}=360a_2+120a_3+30a_4+6a_5+a_6+\frac{1}{7}a_7</cmath> | Notice that if we divide everything by <math>7</math> then we would have:<cmath>514+\frac{2}{7}=360a_2+120a_3+30a_4+6a_5+a_6+\frac{1}{7}a_7</cmath> | ||
Since <math>0 \le a_7<7</math> and <math>a_7</math> must be an integer, then we have <math>\frac{2}{7}=\frac{1}{7}a_7</math>, so <math>a_7=2</math>. | Since <math>0 \le a_7<7</math> and <math>a_7</math> must be an integer, then we have <math>\frac{2}{7}=\frac{1}{7}a_7</math>, so <math>a_7=2</math>. | ||
+ | |||
Similarly, if we divide everything by <math>6</math>, then we would have: <cmath>85+\frac{4}{6}=60a_2+20a_3+5a_4+a_5+\frac{1}{6}a_6</cmath> | Similarly, if we divide everything by <math>6</math>, then we would have: <cmath>85+\frac{4}{6}=60a_2+20a_3+5a_4+a_5+\frac{1}{6}a_6</cmath> | ||
Again, since <math>0 \le a_6<6</math> and <math>a_6</math> must be an integer, we have <math>\frac{4}{6}=\frac{1}{6}a_6</math>, so <math>a_6=4</math>. | Again, since <math>0 \le a_6<6</math> and <math>a_6</math> must be an integer, we have <math>\frac{4}{6}=\frac{1}{6}a_6</math>, so <math>a_6=4</math>. |
Revision as of 11:01, 10 July 2023
Contents
Problem
There are unique integers such that
where for . Find .
Solution 1(Modular Functions)
Multiply out the to get
By Wilson's Theorem (or by straightforward division), , so . Then we move to the left and divide through by to obtain
We then repeat this procedure , from which it follows that , and so forth. Continuing, we find the unique solution to be (uniqueness is assured by the Division Theorem). The answer is .
Solution 2(Basic Algebra and Bashing)
We start by multiplying both sides by , and we get: After doing some guess and check, we find that the answer is .
~aopspandy
Solution 3 (The Easiest and Most Intuitive)
Let's clear up the fractions: Notice that if we divide everything by then we would have: Since and must be an integer, then we have , so .
Similarly, if we divide everything by , then we would have: Again, since and must be an integer, we have , so .
The pattern repeats itself, so in the end we have , , , , , . So
~BurpSuite, with a help from ostriches88
See also
1999 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 24 |
Followed by Problem 26 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 | ||
All AHSME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.