Difference between revisions of "1979 AHSME Problems/Problem 22"
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==Solution== | ==Solution== | ||
− | + | The equation is equivalent to <math>m(m+1)(m+5) = 27n^3 + 27n^2 + 9n + 1</math>. Taking mod 3, we get | |
− | + | <math>m(m+1)(m+2)=1 (\bmod 3)</math>. However, <math>m(m+1)(m+2)</math> is always divisible by <math>3</math> for any integer <math>m</math>. Thus, the answer is <math>\boxed{\textbf{(A)} 0}</math> | |
− | + | Solution by mickyboy789 | |
== See also == | == See also == |
Latest revision as of 10:26, 19 August 2021
Problem 22
Find the number of pairs of integers which satisfy the equation .
Solution
The equation is equivalent to . Taking mod 3, we get . However, is always divisible by for any integer . Thus, the answer is Solution by mickyboy789
See also
1979 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 21 |
Followed by Problem 23 | |
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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