What is the remainder when a is divided by 4?

Positive integer \(a\) divided by positive integer \(d\) yields a reminder of \(r\) can always be expressed as \(a=qd+r\), where \(q\) is called a quotient and \(r\) is called a remainder, note here that \(0\leq{r}<d\) (remainder is non-negative integer and always less than divisor).

So according to above, when positive integer \(a\) is less than divisor \(d\) then remainder upon division \(a\) by \(d\) is always equals to \(a\), for example 5 divided by 10 yields reminder of 5. So when 1 is divided by 4 remainder is 1.

Or algebraically: 1 divided by 4 can be expressed as \(1=0*4+1\), so \(r=1\).

Back to the original question:

What is the remainder when a is divided by 4?

(1) a is the square of an odd integer --> \(a=(2k+1)^2=4k^2+4k+1\), first two terms (4k^2 and 4k) are divisible by 4 and the third term (1) when divided by 4 yields the remainder of 1. Sufficient.

Or you can try several numbers for \(a\):
\(a=1^1=1\) --> 1 divided by 4 yields remainder of 1;
\(a=3^1=9\) --> 9 divided by 4 yields remainder of 1;
\(a=5^1=25\) --> 25 divided by 4 yields remainder of 1;
...

(2) a is a multiple of 3 --> clearly insufficient as \(a\) can as well be a multiple of 4, 12 for example, and in this case the remainder will be 0, and it also can not be a multiple of 4, 3 for example, and in this case the remainder will be 3. Not sufficient.

Answer: A.

Questions on remainders:

PS:
remainder-101074.html
remainder-problem-92629.html
number-properties-question-from-qr-2nd-edition-ps-96030.html
remainder-when-k-96127.html
ps-0-to-50-inclusive-remainder-76984.html
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remainder-problem-88102.html

DS:
remainder-problem-101740.html
remainder-101663.html
ds-gcd-of-numbers-101360.html
data-sufficiency-with-remainder-98529.html
sum-of-remainders-99943.html
ds8-93971.html
need-solution-98567.html
gmat-prep-ds-remainder-96366.html
gmat-prep-ds-93364.html
ds-from-gmatprep-96712.html
remainder-problem-divisible-by-86839.html
gmat-prep-2-remainder-86155.html
remainder-94472.html
remainder-problem-84967.html

Hope it helps.