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How positive integers less than 100 have exactly 4 odd

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GMAT Instructor
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How positive integers less than 100 have exactly 4 odd [#permalink] New post 11 Jul 2006, 01:45
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E

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How positive integers less than 100 have exactly 4 odd factors but no even factors?

(A) 13 (B) 14 (C) 15 (D) 16 (E) 17

(Please don't look at all positive integers less than 100 one by one)
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 [#permalink] New post 11 Jul 2006, 03:57
When integers in each of these triplets are multiplied we will get numbers less than 100 with 4 ODD factors each
1,3,5 1,5,7 1,7,11
1,3,9 1,5,11 1,7,13
1,3,7 1,5,13
1,3,11 1,5,17
1,3,13 1,5,19
1,3,17
1,3,19
1,3,23
1,3,29
1,3,31
I will try to explain my logic here. If all factors are odd then the numbers themselves are odd. So we narrow down the scope to 50 numbers. The number 1 will present in each and the number itself is also a factor. So we need 2 more factors. When they are primes then we need no more. The only exception is 27. Starting from the LEAST ODD PRIME(3) think that the ans is E
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 [#permalink] New post 11 Jul 2006, 06:36
Excellent! As you point out, an integer has exactly four factors if and only if it is the product of two different prime numbers or the cube of a prime number.

Looking at the odd prime numbers {3,5,7,11,13,17,19,23,29,31...}

For a product less than 100

Case I 3 multiplied by any of {5,7,11,13...31}....... 9 numbers
Case II 5 "............................ " {7,...,19}................. 5 numbers
Case III 7 " ..............................." {11,13} ............... 2 numbers

Case IV 3*3*3=27 ............................................... 1 number

TOTAL: 17
OA: E

Was this a doable question in less than 3 or 4 minutes?

Last edited by kevincan on 11 Jul 2006, 06:42, edited 1 time in total.
  [#permalink] 11 Jul 2006, 06:36
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