If 97-79 is divided by 64 - then the remainder isA- 1B- 11C- 0D- 61

The correct option is C 0Given 9^7+7^9 =(1+8)^7 (1 8)^9 =[1+ ^7C_18+…+^7C_78^7] [1 ^9C_18+… ^9C_98^9 ] =8[ ^7C_1+ ^9C_1]+64k, k ∈ℤ nbsp;= 128+64k = 64 ...

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Byju's Answer

Standard XII

Mathematics

Finding Integral Part of Numbers of the Form a^b Where a Is Irrational

If 97+79 is d...

Question

If $9^{7} + 7^{9}$ is divided by $64$ , then the remainder is

1

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Solution

The correct option is B $0$
Given
$9^{7} + 7^{9} = (1 + 8)^{7} - (1 - 8)^{9} = [1 +^{7} C_{1} 8 + \dots +^{7} C_{7} 8^{7}] - [1 -^{9} C_{1} 8 + \dots -^{9} C_{9} 8^{9}] = 8 [^{7} C_{1} +^{9} C_{1}] + 64 k, k \in Z = 128 + 64 k = 64 (k + 2)$

Therefore, the given number is divisible by $64$ , so remainder is $0$ .

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Finding Integral Part of Numbers of the Form a^b Where a Is Irrational

Standard XII Mathematics

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If 97+79 is divided by 64, then the remainder is

The correct option is B 0Given 97+79=(1+8)7−(1−8)9=[1+7C18+…+7C787]−[1−9C18+…−9C989]=8[ 7C1+ 9C1]+64k, k∈Z=128+64k=64(k+2) Therefore, the given number is divisible by 64, so remainder is 0.

If $9^{7} + 7^{9}$ is divided by $64$ , then the remainder is