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IMO EFA,(E)

1st A
2nd B
.....
.....

Periodicity is 6;

nmod 6 will give the digit;

Here n=713; According to the pattern, it is the 5th repetitive element, which is E.
So EFA.
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Six boxes: ABCEDF

Since each box letter is repeated in a periodicity of 6, we know that 714th box will be F, since 714 is divisible by 6.

From there we count backwards one to count the 713th box as letter E.

713th, 714th, and 715th box as follows: EFA

Option E

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