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Hadrienlbb
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If a is a positive integer and 25 Nov 2017, 15:33
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If a is a positive integer and \(\frac{22a}{210}\) is also an integer, how many different positive prime factors could a possibly have at most given that a is less than 150?

A) Two
B) Three
C) Four
D) Five
E) Six
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B
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If a is a positive integer and 25 Nov 2017, 20:37
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Hadrienlbb
If a is a positive integer and \(\frac{22a}{210}\) is also an integer, how many different positive prime factors could a possibly have at most given that a is less than 150?

A) Two
B) Three
C) Four
D) Five
E) Six

Ans : B

210 = 2*3*5*7
22a = 2*11*a

For 22a to be perfectly divisible by 210, a should have the missing prime factors 3,5 and 7.
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If a is a positive integer and 26 Nov 2017, 01:44
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Hadrienlbb
If a is a positive integer and \(\frac{22a}{210}\) is also an integer, how many different positive prime factors could a possibly have at most given that a is less than 150?

A) Two
B) Three
C) Four
D) Five
E) Six

Similar question from GMAT Prep: https://gmatclub.com/forum/in-n-is-a-po ... 00763.html
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If a is a positive integer and 26 Nov 2017, 18:42
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210 = 2x3x5x7
22= 2x11
22a/210 = 11a/3x5x7 is a integer

a must contain 3,5,7
prime factor a= kx3x5x7=kx105 if k=2 therefore a>150
so three prime factors
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If a is a positive integer and 29 Nov 2017, 10:59
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Hadrienlbb
If a is a positive integer and \(\frac{22a}{210}\) is also an integer, how many different positive prime factors could a possibly have at most given that a is less than 150?

A) Two
B) Three
C) Four
D) Five
E) Six

Let’s simplify the given expression:

22a/210 = 11a/105

Thus, we see that a/105 is an integer. In other words, a = 105k for some positive integer k. However, since we are given that a is less than 150, we see that k must be 1 and thus a must be 105.

Since 105 = 7 x 3 x 5, we see that a could have 3 different prime factors.

Answer: B
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If a is a positive integer and 07 Jan 2025, 08:37
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