The remainder when 7^{7^{7^{...7}}} (22 times | vedclass

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(B) Let $x = 7^{7^{...7}}$ ($22$ times $7$). We want to find $x \pmod{48}$.Note that $7^2 = 49 \equiv 1 \pmod{48}$.Since the exponent of $7$ is $7^{7^

Vedclass · Accepted Answer

$7$

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(B) Let $x = 7^{7^{...7}}$ ($22$ times $7$). We want to find $x \pmod{48}$.Note that $7^2 = 49 \equiv 1 \pmod{48}$.Since the exponent of $7$ is $7^{7^{...7}}$ ($21$ times $7$), which is clearly an odd number, let the exponent be $2k+1$ for some integer $k$.Then $7^{2k+1} = 7 	imes (7^2)^k = 7 	imes (49)^k$.Since $49 \equiv 1 \pmod{48}$, we have $49^k \equiv 1^k \equiv 1 \pmod{48}$.Therefore, $7 	imes (49)^k \equiv 7 	imes 1 \equiv 7 \pmod{48}$.The remainder is $7$.

The remainder when $7^{7^{7^{...7}}}$ ($22$ times $7$) is divided by $48$ is

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