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JEE Main · 2025 · 08 Apr 2025 · Shift 2 · Binomial Theorem

The product of the last two digits of

(1919)^{1919}

is

Answer: 63

Solution

Only the last two digits matter, so work modulo

100

. As

1919 \equiv 19

, evaluate

1 9^{1919} mod 100

. Here

1 9^{2} = 361 \equiv 61

and

1 9^{4} \equiv 6 1^{2} = 3721 \equiv 21

, so

1 9^{5} \equiv 21 \cdot 19 = 399 \equiv - 1

, giving

1 9^{10} \equiv 1

. Since

1919 \equiv 9 (mod 10)

,

1 9^{1919} \equiv 1 9^{9} = 1 9^{5} \cdot 1 9^{4} \equiv (- 1) (21) = - 21 \equiv 79 (mod 100)

. The last two digits are

7

and

9

, whose product is

63

.

JEE Main · 2025 · 08 Apr 2025 · Shift 2 · Integer answer

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