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STD 11 - 7. binomial theoram — Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceMathsSTD 11 - 7. binomial theoram1 Mark
MCQ
The remainder when $(2021)^{2022}+(2022)^{ 2021 }$ is divided by $7$ is.
  • ✓
    $0$
  • B
    $1$
  • C
    $2$
  • D
    $6$
✓

Answer

Correct option: A.
$0$
a
$(2021)^{\text {a022 }}+(2022)^{2011}$

$=(2023-2)^{\text {an2 }}+(2023-1)^{3031}$

$=7 n_{1}+2^{m 2 n 2}+7 n _{2}-1$

$=7\left( n _{1}+ n _{2}\right)+8^{674}-1$

$=7\left(n_{1}+n_{2}\right)+(7-1)^{674}-1$

$=7\left(n_{1}+n_{2}\right)+7 n_{3}+1-1$

$=7\left(n_{1}+n_{2}+n_{3}\right)$

$\therefore$ Given number is divisible by $7$ hence remainder is zero

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