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4t+3 would leave the same remainder as 4t-1.
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Great responses in the thread.

I took a slightly different route.

n/4 results in remainder of 3. n=4k+3

if k = 0, n = 3
if k = 1, n = 7
if k = 2, n = 11...

Resulting in possible n values of 3,7,11,15,19,23

I then looked at the answer choices which all have 4t. With this multiple of 4 I plugged in 3 for t to get 12. Resulting in:

A) 14 (x)
B) 12 (x)
C) 11 (correct)
D) 10 (x)
E) 9 (x)

Of this, only 11 is on our list. Answer C.
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