JEE Main 2025 (Online) 8th April Evening ShiftQuestion 1
The product of the last two digits of $(1919)^{1919}$ is $\_\_\_\_\_\_$ .
Binomial Theorem · Remainder and Divisibility
Answer$63$
Solution
Concept. Only the last two digits matter, so work modulo $100$ and find a power of $19$ that is $\pm1$. Formulas used. (1) $a\equiv b\pmod{100}\Rightarrow a^n\equiv b^n\pmod{100}$Step 1 - Reduce the base. $1919\equiv19\pmod{100}$, so $1919^{1919}\equiv19^{1919}$. Step 2 - Small powers of $19$. $$19^2=361\equiv61,\qquad 19^4\equiv61^2=3721\equiv21,$$ $$19^5\equiv21\cdot19=399\equiv-1\pmod{100}.$$ Step 3 - Split the exponent. $1919=5\cdot383+4$, so $$19^{1919}=\left(19^5\right)^{383}\cdot19^4\equiv(-1)^{383}\cdot21=-21\equiv79\pmod{100}.$$ Step 4 - Product of digits. The last two digits are $7$ and $9$: $7\times9=63$. Answer: 63