JEE MAIN - Mathematics (2023 - 31st January Morning Shift - No. 16)
The remainder on dividing $$5^{99}$$ by 11 is ____________.
Answer
9
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 $=9$
$=625\left[5^{5}\right]^{19}$
$=625[3125]^{19}$
$=625[3124+1]^{19}$
$=625[11 \mathrm{k} \times 19+1]$
$=625 \times 11 \mathrm{k} \times 19+625$
$=11 \mathrm{k}_{1}+616+9$
$=11\left(\mathrm{k}_{2}\right)+9$
Remainder $=9$
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