JEE MAIN - Mathematics (2023 - 1st February Morning Shift - No. 17)
The remainder, when $$19^{200}+23^{200}$$ is divided by 49 , is ___________.
Answer
29
Explanation
$19^{200}+23^{200}$
= $(21-2)^{200}+(21+2)^{200}=49 \lambda+2^{201}$
Now, $2^{201}=8^{67}=(7+1)^{67}=49 \lambda+7 \times 67+1$
$=49 \lambda+470$
$=49(\lambda+9)+29$
$$ \therefore $$ Remainder $=29$
= $(21-2)^{200}+(21+2)^{200}=49 \lambda+2^{201}$
Now, $2^{201}=8^{67}=(7+1)^{67}=49 \lambda+7 \times 67+1$
$=49 \lambda+470$
$=49(\lambda+9)+29$
$$ \therefore $$ Remainder $=29$
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