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If t is divisible by 12, what is the least possible integer
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Updated on: 16 Oct 2018, 23:04
Updated on: 16 Oct 2018, 23:04
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If t is divisible by 12, what is the least possible integer value of a for which \(\frac{t^2}{2^a}\) might not be an integer?
(A) 2
(B) 3
(C) 4
(D) 5
(E) 6
(A) 2
(B) 3
(C) 4
(D) 5
(E) 6
Originally posted by ruposh6240 on 16 Oct 2018, 06:06.
Last edited by Bunuel on 16 Oct 2018, 23:04, edited 2 times in total.
Last edited by Bunuel on 16 Oct 2018, 23:04, edited 2 times in total.
Edited by Carcass
Re: If t is divisible by 12, what is the least possible integer
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21 Sep 2019, 10:27
21 Sep 2019, 10:27
1
t is divisible by 12 -> t= 12*X1*X2*..Xn
t=(2^2)*3*X1*X2*..Xn
t^2=2^4*3^2*X1^2*X2^2*..Xn^2
for a=2,3 and 4, t^2 is divisible but for a=5 and 6, t^2 might not be divisible.
Hence least value of a for which t^2 might not be divisible is 5
t=(2^2)*3*X1*X2*..Xn
t^2=2^4*3^2*X1^2*X2^2*..Xn^2
for a=2,3 and 4, t^2 is divisible but for a=5 and 6, t^2 might not be divisible.
Hence least value of a for which t^2 might not be divisible is 5
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Re: If t is divisible by 12, what is the least possible integer
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19 Sep 2022, 07:45
19 Sep 2022, 07:45
1
Given that t is divisible by 12 and we need to find what is the least possible integer value of a for which \(\frac{t^2}{2^a}\) might not be an integer
t is divisible by 12 => t is a multiple of 12 => t = 12x where x is an integer
\(\frac{t^2}{2^a}\) = \(\frac{(12x)^2}{2^a}\) = \(\frac{(2^2*3*x)^2}{2^a}\) = \(\frac{2^4*3^2*x^2}{2^a}\)
Now, in numerator 2 has a power of 4. So, a should be more than 4 for \(\frac{t^2}{2^a}\) to not be an integer
=> a = 5
So, Answer will be D.
Hope it helps!
t is divisible by 12 => t is a multiple of 12 => t = 12x where x is an integer
\(\frac{t^2}{2^a}\) = \(\frac{(12x)^2}{2^a}\) = \(\frac{(2^2*3*x)^2}{2^a}\) = \(\frac{2^4*3^2*x^2}{2^a}\)
Now, in numerator 2 has a power of 4. So, a should be more than 4 for \(\frac{t^2}{2^a}\) to not be an integer
=> a = 5
So, Answer will be D.
Hope it helps!
