Difference between revisions of "1979 AHSME Problems/Problem 22"
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Take equation is equivalent to <math>m(m+1)(m+5) = 27n^3 + 27n^2 + 9n + 1</math>. Taking mod 3, we get | Take 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 /mod 3</math> | + | <math>m(m+1)(m+2)=1 (mod 3)</math> or <math>0=1 (mod 3)</math> which is a contradiction. Thus, the answer is <math>(A) 0</math> |
== See also == | == See also == |
Revision as of 00:25, 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 .
Solution 2:
Take equation is equivalent to . Taking mod 3, we get
or which is a contradiction. Thus, the answer is
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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