JEE Main 2023 (Online) 8th April Evening ShiftQuestion 1
$25^{190}-19^{190}-8^{190}+2^{190}$ is divisible by
Binomial Theorem · Remainder and Divisibility
Solution
Concept. Pair $25$ with $8$ and $19$ with $2$: both differences are $17$. Formulas used. (1) $(a-b)\mid\left(a^n-b^n\right)$(2) $a^6\equiv1\pmod7$ for $7\nmid a$Step 1 - Divisible by $17$. Group the terms: $$\left(25^{190}-8^{190}\right)-\left(19^{190}-2^{190}\right).$$ By (1), each bracket is divisible by $25-8=17$ and $19-2=17$. Step 2 - Divisible by $2$. odd $-$ odd $-$ even $+$ even is even. So the number is divisible by $34$. Step 3 - Modulo $7$. $25\equiv4$, $19\equiv5$, $8\equiv1$, $2\equiv2$, and $190=6\cdot31+4$. By (2), $$4^4-5^4-1+2^4\equiv4-2-1+2=3\pmod7.$$ So the number is not divisible by $7$, hence not by $14$. Answer: (C)