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Number properties

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Intern
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Number properties [#permalink] New post 03 Apr 2006, 19:00
Hey guys...I was wondering what the best shortcut for this question is...


n= (2*3*5*7*11*13)/(77k)

If n is an integer and then which of the following could be the value of k?
(A) 22
(B) 26
(C) 35
(D) 54
(E) 60

Answer:B
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 [#permalink] New post 03 Apr 2006, 19:23
n=(2*3*5*13)/k = 390/k

390mod26 = 0
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 [#permalink] New post 03 Apr 2006, 19:24
You start with n=(2*3*5*7*11*13)/77k

-cancel out 11 & 7 => n=(2*3*5*13)/k
=> nk=2*3*5*13

-Now look at your answer choices and their prime factorization:

A 11*2
B 13*2
C 5*7
D 9*3*2
E 2*2*3*15

Since the question specifically says that n is an integer, the answer choice must contain the prime 13, therefore the answer is B.
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 [#permalink] New post 03 Apr 2006, 19:59
Ok...I didn't carry out the last step...it is perfectly clear now..thanks for your help!
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 [#permalink] New post 03 Apr 2006, 20:20
trublu wrote:
390mod26 = 0
:roll:
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 [#permalink] New post 04 Apr 2006, 19:24
Do a prime factorization of 77 to get 11 * 7

We're told n is an integer, so k must be cancel out 2,3,5,13 or a combination of any of the 4.

A) 22 - 11*2 --> out
B) 26 - 2*13 --> possible
C) 35 - 7*5 --> out
D) 54 - 3*3*3*2 --> out
E) 60 - 2*5*2*3 --> out

B is the only choice that works.
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 [#permalink] New post 04 Apr 2006, 20:14
best short cut

n= (2*3*5*7*11*13)/(77k)

n = (2*3*5*7*11*13)/(7*11*k)

n = (2*3*5*13)/(k)

n = 3*13*5*2/k

22 = 11 * 2
26 = 2 * 13
35 = 7 * 5
54 = 2*3*3*3
60 = 5*2*3*2

26 has all digits :)
  [#permalink] 04 Apr 2006, 20:14
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