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JEE Main · 2023 · 01 Feb 2023 · Shift 1 · Binomial Theorem

The remainder when

1 9^{200} + 2 3^{200}

is divided by 49 , is ______.

Answer: 29

Solution

Writing

19 = 21 - 2

and

23 = 21 + 2

, the sum

(21 - 2)^{200} + (21 + 2)^{200}

keeps only the even-power terms,

2 \sum (2 k 200) 2 1^{200 - 2 k} 2^{2 k}

. Every term carrying

2 1^{2}

or a higher power is divisible by

49

, leaving only

2 \cdot 2^{200} = 2^{201}

to reduce. Since

2^{201} = 8^{67} = (1 + 7)^{67} \equiv 1 + 67 \cdot 7 = 470 \equiv 29 (mod 49)

, the remainder is

29

.

JEE Main · 2023 · 01 Feb 2023 · Shift 1 · Integer answer

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