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What is the last digit in the decimal representation of $7^{19522}$?
in Quantitative Aptitude recategorized by
469 views

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An excellent article on solving these types of problems. 
https://brilliant.org/wiki/finding-the-last-digit-of-a-power/

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1 vote
1 vote
$7^2\equiv 9\; \mod \; 10$

$7^3 \equiv 7\ast 7^2\equiv 7\ast 9 \equiv 3 \; \mod \; 10$
and finally

$7^4\equiv 7\ast 7^3 \equiv 7 \ast 3 \equiv 1 \; \mod \; 10$

$7^{19522} = (7^4)^{4880} \times 7^2$

$7^{19522} \equiv 7^2 \equiv 9\; (\mod\; 10)$
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$7^{^{19522}} mod 10 = (7^{^{2}})^{9761} mod 10 = (49)^{^{9761}}mod 10=9^{^{9761}}mod10=-1^{9761}mod10=-1 mod 10=9$

So we can say last digit is 9.
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