Hi everyone, I wanted some help to see if there is a faster way to solve this question:
" if t is divisible by 12, what is the least possible integer value of a for which (t^2/2^a) might not be an integer "
A. 2
B.3
C.4
D.5
E.6
I chose 12 for which makes t^2= 144. and then i just did 144/ 2^a and plugged in each answer choice for A until I found the lowest number that would give me a non integer. However, I wanted to see if there was another way to solve this ?
You’re thinking it the right direction :
If t is divisible by 12, then t must have at least the factors of 12, which are 2^2 * 3 . When we square t , the powers of its prime factors also get squared. So, t^2 will have at least 2^4 * 3^2(144) as factors.
Now, we want to find the smallest a such that \dfrac{t^2}{2^a} might not be an integer. This will happen when we try to divide t^2 by more 2’s than it has. Since t^2 has four 2’s (from 2^4 ) , if we try to divide by more than four 2’s (i.e., 2^5 ), \dfrac{t^2}{2^a} might not be an integer.