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Posts: 21
Updated on: 11 Jun 2023, 01:14
Originally posted by shikha_0203 on 11 Jun 2023, 01:12.
Last edited by Bunuel on 11 Jun 2023, 01:14, edited 1 time in total.
Last edited by Bunuel on 11 Jun 2023, 01:14, edited 1 time in total.
Edited the question and the tags.
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If a and p are integers in the equation a^2 = 12p, then p must be divisible by which of the following?
A. 2
B. 3
C. 6
D. 12
E. 24
A. 2
B. 3
C. 6
D. 12
E. 24
Posts: 23
a^2 = 12p.
a^2/12 = p. Since a and p are integers, a^2 must be divisible by 12.
Take a^2 = 36. a^2/12 = 3 = p. There is no perfect square smaller than 36 which is divisible by 12. Therefore, p must be divisible by 3.
a^2/12 = p. Since a and p are integers, a^2 must be divisible by 12.
Take a^2 = 36. a^2/12 = 3 = p. There is no perfect square smaller than 36 which is divisible by 12. Therefore, p must be divisible by 3.
If a and p are integers in the equation a^2 = 12p, then p must be divisible by which of the following?
A. 2
B. 3
C. 6
D. 12
E. 24
a^2 = 12p = 2^2 * 3 * p
Since there's only one 3 along 2^2 when 12p needs to be a square. The least p can be is 3 or a multiple of 3 that is square of an integer in itself.
Example: 3 or 3 * 2^2 or 3 * 3^3 and so on.
Hence, p must be a multiple of 3.
Answer B.
_________________
A. 2
B. 3
C. 6
D. 12
E. 24
a^2 = 12p = 2^2 * 3 * p
Since there's only one 3 along 2^2 when 12p needs to be a square. The least p can be is 3 or a multiple of 3 that is square of an integer in itself.
Example: 3 or 3 * 2^2 or 3 * 3^3 and so on.
Hence, p must be a multiple of 3.
Answer B.
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Expert reply
Posts: 2,973
shikha_0203
If a = 6, then p = (6^2)/12 = 3. In this case, the only answer choice that divides evenly into p is 3.
Answer: B
(Note: If we can eliminate four answer choices, then the fifth choice must be the correct answer.)
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