Difference between revisions of "1979 AHSME Problems/Problem 22"
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The equation is equivalent to <math>m(m+1)(m+5) = 27n^3 + 27n^2 + 9n + 1</math>. Taking mod 3, we get | 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 ( | + | <math>m(m+1)(m+2)=1 (\bmod 3)</math> or <math>0=1 (\bmod 3)</math> which is a contradiction. Thus, the answer is <math>(A) 0</math> |
Solution by mickyboy789 | Solution by mickyboy789 | ||
Revision as of 01:27, 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:
The equation is equivalent to . Taking mod 3, we get
or
which is a contradiction. 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.