Difference between revisions of "2011 AMC 12B Problems/Problem 15"
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<math>2^{24}-1 = (2^{12} +1) * 5 * 13 * 3^2 * 7</math> | <math>2^{24}-1 = (2^{12} +1) * 5 * 13 * 3^2 * 7</math> | ||
− | + | Applying sum of cubes: | |
<math>2^{12}+1 = (2^4 + 1)(2^8 - 2^4 + 1) </math> | <math>2^{12}+1 = (2^4 + 1)(2^8 - 2^4 + 1) </math> |
Revision as of 21:57, 5 January 2013
Problem 15
How many positive two-digits integers are factors of ?
Solution
From repeated application of difference of squares:
Applying sum of cubes:
A quick check shows is prime. Thus, the only factors to be concerned about are
, since multiplying by
will make any factor too large.
Multiply by
or
will give a two digit factor;
itself will also work. The next smallest factor,
, gives a three digit number. Thus, there are
factors which are multiples of
.
Multiply by
or
will also give a two digit factor, as well as
itself. Higher numbers will not work, giving an additional
factors.
Multiply by
or
for a two digit factor. There are no mare factors to check, as all factors which include
are already counted. Thus, there are an additional
factors.
Multiply by
or
for a two digit factor. All higher factors have been counted already, so there are
more factors.
Thus, the total number of factors is
See also
2011 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 14 |
Followed by Problem 16 |
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 | |
All AMC 12 Problems and Solutions |