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The remainder on dividing 599 by 11 is | (Mathematics) | DivineJEE

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Mathematics

Exam

Mathematics

The remainder on dividing $5^{99}$ by 11 is ____________.

Explanation

$5^{99}=5^{4} .5^{95}$

$=625\left[5^{5}\right]^{19}$

$=625[3125]^{19}$

$=625[3124+1]^{19}$

$=625[11 \mathrm{k} \times 19+1]$

$=625 \times 11 \mathrm{k} \times 19+625$

$=11 \mathrm{k}_{1}+616+9$

$=11\left(\mathrm{k}_{2}\right)+9$

Remainder $=9$

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