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Bunuel
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If u is a multiple of prime number is u a multiple of v^2? Updated on: 06 Oct 2020, 21:46
Originally posted by Bunuel on 29 Sep 2020, 02:40.
Last edited by chetan2u on 06 Oct 2020, 21:46, edited 1 time in total.
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21% (01:34) correct 79% (01:44) wrong based on 43 sessions

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If u is a multiple of prime number v, is u a multiple of v^2?

(1) v < 6

(2) u = 42

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B
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If u is a multiple of prime number is u a multiple of v^2? 29 Sep 2020, 12:32
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Given: If u is a multiple of prime number
To find: is u a multiple of v^2?

(1) v < 6 (V can be a prime no. or can not be a prime number)
Case 1: V = 5 (prime no.)
V^2 = 5^2 = 25 (u is also a multiple of v^2)

Case 2: V = 4 (non-prime no.)
V^2 = 4^2 = 16 (u is not a multiple of v^2)
(Insufficient)

(2) u = 42
u = 2*3*7 (multiple of prime no.)
therefore, u is not a multiple of any square no.
so u is not a multiple of V^2
(sufficient)

Answer is B
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If u is a multiple of prime number is u a multiple of v^2? 29 Sep 2020, 13:35
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yashikaaggarwal
Given: If u is a multiple of prime number
To find: is u a multiple of v^2?

(1) v < 6 (V can be a prime no. or can not be a prime number)
Case 1: V = 5 (prime no.)
V^2 = 5^2 = 25 (u is also a multiple of v^2)

Case 2: V = 4 (non-prime no.)
V^2 = 4^2 = 16 (u is not a multiple of v^2)
(Insufficient)

(2) u = 42
u = 2*3*7 (multiple of prime no.)
therefore, u is not a multiple of any square no.
so u is not a multiple of V^2
(sufficient)

Answer is B
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can't it be multiple of sqrt 2?, as it is not given that v is a integer
or v can be 1 also
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If u is a multiple of prime number is u a multiple of v^2? 29 Sep 2020, 13:44
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bhupendersingh27

can't it be multiple of sqrt 2?, as it is not given that v is a integer
or v can be 1 also
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Yeah, V can be fraction as well as non-integer.
I just took two cases to prove statement 1 Insufficient
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If u is a multiple of prime number is u a multiple of v^2? 30 Sep 2020, 08:58
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Bunuel
If u is a multiple of prime number is u a multiple of v^2?

(1) v < 6

(2) u = 42
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I believe the question is supposed to be "If v is a multiple of a prime number, is u a multiple of \(v^2\)?"

"v is a multiple of a prime number" really just means v cannot be 0 or 1, and it will be an integer that is greater than or equal to 2.

Now let's look at the second statement, with \(u = 42 = 6*7 = 2*3*7\) it does not have any repeating factors to fit in a square that is not 1. Then \(v^2\) cannot be a factor of u, thus statement 2 alone is sufficient and we'd choose B.
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If u is a multiple of prime number is u a multiple of v^2? 02 Oct 2020, 06:43
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yashikaaggarwal
Given: If u is a multiple of prime number
To find: is u a multiple of v^2?

(1) v < 6 (V can be a prime no. or can not be a prime number)
Case 1: V = 5 (prime no.)
V^2 = 5^2 = 25 (u is also a multiple of v^2)

Case 2: V = 4 (non-prime no.)
V^2 = 4^2 = 16 (u is not a multiple of v^2)
(Insufficient)

(2) u = 42
u = 2*3*7 (multiple of prime no.)
therefore, u is not a multiple of any square no.
so u is not a multiple of V^2


Hi,

In the second statement if U=42 ,then U can be multiple Of V^2 if V=1.
If It was given the V is multiple of prime number than it stands valid that 42 will not be the multiple of any square number.
Since, there is no restriction on V then we can consider V=1.
In that case B will not be the answer
(sufficient)

Answer is B
Show more
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If u is a multiple of prime number is u a multiple of v^2? 04 Oct 2020, 22:56
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Bunuel
If u is a multiple of prime number is u a multiple of v^2?

(1) v < 6

(2) u = 42
Show more

u is a multiple of prime number tells us
a) u is a positive integer
b) u >1

Let u=ax, where x is a prime number, and a is a positive integer.

Is u a multiple of v^2?
Nothing known about v?

(1) v < 6..
u and v could be anything..
u=25, v=5..Yes
u=4, v=5..No
Insuff

(2) u = 42
u=42=2*3*7..
If \(v=\sqrt{2}=\sqrt{3}=\sqrt{7}\), answer is YES
If v is an integer, answer is NO
Insuff

Combined..
If \(v=\sqrt{2}..or..\sqrt{3}..or..\sqrt{7}\), answer is YES
If v is an integer, (2, 3, 7, 4) answer is NO

E

Bunuel, Either there is a typo or the answer requires to be changed.

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