Answer: B) 4
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Step-by-step explanation:
"n^4 is divisible by 32" means 32k = n^4 for some integer k.
In other words, 32 is a factor of n^4.
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Do a bit of algebraic transformations to get the following
32k = n^4
16*2k = n^4
2^4*2k = n^4
2k = (n^4)/(2^4)
2k = (n/2)^4
This shows that (n/2)^4 is an even number.
Consequently, (n/2)^2 must also be even. If it were odd, then we'd have odd*odd = odd which would be contradictory to (n/2)^4 being even. Furthermore, this means n/2 is even as well.
The logic chain is
(n/2)^4 is even --> (n/2)^2 is even --> n/2 is even
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Because n/2 is an even integer, this means
n/2 = 2m for some integer m
Multiply both sides by 2
n/2 = 2m
2*n/2 = 2*2m
n = 4m .... we'll use this later
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Now let's divide n over 32 and seek out the remainder r. Let q be the quotient.
n/32 = q + r/32
The quotient is the whole part and the remainder is the fractional part leftover.
Multiply both sides by 32 so we clear out the fractions
n/32 = q + r/32
32(n/32) = 32(q + r/32)
32(n/32) = 32(q) + 32(r/32)
n = 32q + r
So n is some multiple of 32 plus the remainder r.
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Solve for r and plug in n = 4m
n = 32q + r
32q + r = n
r + 32q = n
r = n-32q
r = 4m-32q ... plug in n = 4m
r = 4(m-8q) ... factor out the GCF
r = 4p ... where p = m-8q is an integer
We see that r is a multiple of 4
Therefore, the only possible remainders are 0, 4, 8, 12, 16, 20, 24, and 28
note: after we get to r = 32, the remainder becomes 0 and the pattern of (0,4,8,...) repeats again.
Of the answer choices listed, only choice B is a multiple of 4. So this is why choice B is the answer.