Difference between revisions of "1999 AHSME Problems/Problem 25"
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<cmath>\frac{5 \cdot 6!-2}{7} = 514 = 360a_2 + 120a_3 + 30a_4 + 6a_5 + a_6.</cmath> | <cmath>\frac{5 \cdot 6!-2}{7} = 514 = 360a_2 + 120a_3 + 30a_4 + 6a_5 + a_6.</cmath> | ||
− | We then repeat this procedure <math>\pmod{6}</math>, from which it follows that <math>a_6 \equiv 514 \equiv 4 \pmod{6}</math>, and so forth. Continuing, we find the unique solution to be <math>(a_2, a_3, a_4, a_5, a_6, a_7) = (1,1,1,0,4,2)</math> (uniqueness is assured by the [[Division Theorem]]). The answer is <math>9 \Longrightarrow \mathrm{(B)}</math>. | + | We then repeat this procedure <math>\pmod{6}</math>, from which it follows that <math>a_6 \equiv 514 \equiv 4 \pmod{6}</math>, and so forth. Continuing, we find the unique solution to be <math>(a_2, a_3, a_4, a_5, a_6, a_7) = (1,1,1,0,4,2)</math> (uniqueness is assured by the [[Division Theorem]]). The answer is <math>9 \Longrightarrow \boxed{\mathrm{(B)}}</math>. |
== See also == | == See also == |
Revision as of 16:29, 6 November 2019
Problem
There are unique integers such that
where for . Find .
Solution
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 .
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 |
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