$2^{35}$ divided by $5$ remainders?

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Solution

Find the remainder.
$2^{35}$ divided by $5$ $= 2^{(4 \times 8 + 3)} \div 5$
$= 2^{(4 \times 8)} \times 2^{3} \div 5 = {(16)}^{8} \times 8 \div 5$
$16^{8} \div 5 remainder is 1 [∵ 4294967296 \div 5 = 1] 8 \div 5 remainder is 3$
Remainder $= 1 \times 3 = 3$
$∴$ The remainder is $3$

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