Difference between revisions of "1979 AHSME Problems/Problem 22"
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− | Notice that <math>m^3 + 6m^2 + 5m = 27n^3 + 27n^2 + 9n + 1 = \frac{(3n+1)^3}{27}</math>. Then <math>27(m^3 + 6m^2 + 5m) = (3n+1)^3</math>, and <math>(3n+1)^3 | 27</math>. However, <math>(3n+1)^3</math> will never be divisible by <math>3</math>, nor <math>27</math>, so there are <math>\boxed{\textbf{( | + | Notice that <math>m^3 + 6m^2 + 5m = 27n^3 + 27n^2 + 9n + 1 = \frac{(3n+1)^3}{27}</math>. Then <math>27(m^3 + 6m^2 + 5m) = (3n+1)^3</math>, and <math>(3n+1)^3 | 27</math>. However, <math>(3n+1)^3</math> will never be divisible by <math>3</math>, nor <math>27</math>, so there are <math>\boxed{\textbf{(B) } 10}</math> integer pairs of <math>(m, n)</math>. |
== See also == | == See also == |
Revision as of 00:20, 24 December 2017
Problem 22
Find the number of pairs of integers which satisfy the equation .
Solution
Solution by e_power_pi_times_i
Notice that . Then , and . However, will never be divisible by , nor , so there are integer pairs of .
See also
1979 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 21 |
Followed by Problem 23 | |
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All AHSME Problems and Solutions |
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