Maths Q22 – JEE-MAIN 2025

The product of the last two digits of

(1919)^{1919}

is _______.

Enter Calculated Numerical Value:

Clue 1 of 3: Modular Arithmetic

To find the last two digits of a number, we need to calculate the number modulo 100.

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Step 1: Reduce the base modulo 100✦ Active

To find the last two digits of $(1919)^{1919}$ , we need to calculate $(1919)^{1919} (\mod 100)$ . Since $1919 \equiv 19 (\mod 100)$ , the problem reduces to finding $19^{1919} (\mod 100)$ .

Step 2: Apply Euler's Totient Theorem or Binomial Theorem○ Expand

Using Euler's Totient Theorem, $ϕ (100) = ϕ (2^{2} \cdot 5^{2}) = (2^{2} - 2^{1}) (5^{2} - 5^{1}) = 2 \cdot 20 = 40$ . Since $gcd (19, 100) = 1$ , we have $19^{40} \equiv 1 (\mod 100)$ . We reduce the exponent $1919$ modulo $40$ : $1919 = 40 \times 47 + 39$ . Thus, $19^{1919} \equiv 19^{39} (\mod 100)$ . This can be written as $19^{39} \equiv 19^{40} \cdot 19^{- 1} \equiv 1 \cdot 19^{- 1} (\mod 100)$ . We find the modular inverse of $19 (\mod 100)$ using the Extended Euclidean Algorithm: $19 \times 79 = 1501 \equiv 1 (\mod 100)$ . So, $19^{- 1} \equiv 79 (\mod 100)$ .

Alternatively, using the Binomial Theorem: $19^{1919} = (20 - 1)^{1919}$ . Expanding this, we only need the first two terms for modulo 100: $(20 - 1)^{1919} \equiv (\binom{1919}{0}) (- 1)^{1919} + (\binom{1919}{1}) (20) (- 1)^{1918} (\mod 100)$ . This simplifies to $- 1 + 1919 \times 20 (\mod 100)$ . Since $1919 \equiv 19 (\mod 100)$ , we have $- 1 + 19 \times 20 (\mod 100) = - 1 + 380 (\mod 100) = - 1 + 80 (\mod 100) = 79 (\mod 100)$ .

Step 3: Calculate the product of the last two digits○ Expand

The last two digits of $(1919)^{1919}$ are $7$ and $9$ . The product of these digits is $7 \times 9 = 63$ .

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Maths Question 22 – JEE-MAIN 2025

Clue 1 of 3: Modular Arithmetic

🧠 Full Solution Path