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we can eliminate d 21 which isnt a prime . 97 can be expreesed as 96 + 1
so 97^2 will leave a remainder of 1 so alll that remain are a,b, c . 7 leaves a remainder of 1 again so between 2 and 3 its 3

we can eliminate d 21 which isnt a prime . 97 can be expreesed as 96 + 1 so 97^2 will leave a remainder of 1 so alll that remain are a,b, c . 7 leaves a remainder of 1 again so between 2 and 3 its 3

we can eliminate d 21 which isnt a prime . 97 can be expreesed as 96 + 1 so 97^2 will leave a remainder of 1 so alll that remain are a,b, c . 7 leaves a remainder of 1 again so between 2 and 3 its 3

We have to find LARGEST Prime number that leaves remainder ~= 1

Start from
(E) : 97 is of the form (12n + 1 ) will always give remainder 1
(D) : 21 forget it Not Prime.
(C) : 7 -> 49 Remainder 1
(B) : 3 -> 9 Remainder 9
Thats it : (B) is the answer.

No need to look for (A) because even if that would be answer, 3 will always be greater than 2.

Assuming B is the final ans I will try to give a better explanation.

First off the question is about a prime number so 21 not being a prime is eliminated.

next 97 can be expressed as (96 + 1)^ 2 = 96^2 + 1 + 2*96*1
since (a + b)^2 = a^2 + b^2 + 2*a*b

so since 96 is divisible by 12 97^2/12 leaves a remainder of 1

7^2 / 12 = 49/12 leaves a remainder of 1 so the only 2 options left out are 2 and 3
2^2 / 12 leaves a remainder of 4 and
3^2/ 12 leaves a remainder of 9

but we are looking for the biggest prime number so its 3 between the 2 alternatives