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{(A)} 0 }</math> integer pairs of <math>(m, n)</math>. | 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{(A)} 0 }</math> integer pairs of <math>(m, n)</math>. | ||
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| + | Solution 2: | ||
| + | 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> | ||
== 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
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. 
