The expression 25^{190} - 19^{190} - 8^{190} | vedclass

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Question

(A) Let $E = 25^{190} - 19^{190} - 8^{190} + 2^{190}$.We can rewrite this as $E = (25^{190} - 8^{190}) - (19^{190} - 2^{190})$.Since $a^n - b^n$ is di

Vedclass · Accepted Answer

$34$ but not by $14$

Vedclass · Answer

(A) Let $E = 25^{190} - 19^{190} - 8^{190} + 2^{190}$.We can rewrite this as $E = (25^{190} - 8^{190}) - (19^{190} - 2^{190})$.Since $a^n - b^n$ is divisible by $a - b$ for any even $n$,we have:$25^{190} - 8^{190}$ is divisible by $25 - 8 = 17$.$19^{190} - 2^{190}$ is divisible by $19 - 2 = 17$.Thus,$E$ is divisible by $17$.Also,$E = (25^{190} - 19^{190}) - (8^{190} - 2^{190})$.$25^{190} - 19^{190}$ is divisible by $25 - 19 = 6$.$8^{190} - 2^{190}$ is divisible by $8 - 2 = 6$.Thus,$E$ is divisible by $6$.Since $E$ is divisible by $17$ and $6$,and $	ext{gcd}(17, 6) = 1$,$E$ is divisible by $17 	imes 2 = 34$.Checking for $14$: $25 \equiv 4 \pmod{7}$,$19 \equiv 5 \pmod{7}$,$8 \equiv 1 \pmod{7}$,$2 \equiv 2 \pmod{7}$.$E \equiv 4^{190} - 5^{190} - 1^{190} + 2^{190} \pmod{7}$.Using Fermat's Little Theorem,$a^6 \equiv 1 \pmod{7}$. $190 = 6 	imes 31 + 4$.$E \equiv 4^4 - 5^4 - 1 + 2^4 \equiv 256 - 625 - 1 + 16 \equiv 4 - 2 - 1 + 2 \equiv 3 \pmod{7}$.Since $E 
ot\equiv 0 \pmod{7}$,it is not divisible by $14$.Therefore,it is divisible by $34$ but not by $14$.

The expression $25^{190} - 19^{190} - 8^{190} + 2^{190}$ is divisible by:

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