if 7 9 9 7 is divided by 64 then the remainder is 199030949

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

A. 0

B. 1

C. 2

D. 63

Answer : 0

Solution :
We have,
$$\eqalign{ & {7^9} + {9^7} = {\left( {8 - 1} \right)^9} + {\left( {8 + 1} \right)^7} = {\left( {1 + 8} \right)^7} - {\left( {1 - 8} \right)^9} \cr & = \left[ {1 + {\,^7}{C_1} 8 + {\,^7}{C_2} {8^2} + ..... + {\,^7}{C_7} {8^7}} \right] - \left[ {1 - {\,^9}{C_1} 8 + {\,^9}{C_2} {8^2} - ..... - {\,^9}{C_9} {8^9}} \right] \cr & = {\,^7}{C_1} 8 + {\,^9}{C_1} 8 + \left[ {^7{C_2} + {\,^7}{C_3} 8 + ..... - {\,^9}{C_2} + {\,^9}{C_3} 8 - .....} \right]{8^2} \cr & = 8\left( {7 + 9} \right) + 64k = 8 \cdot 16 + 64k = 64q, \cr} $$
where $$q = k + 2$$
Thus, $$7^9 + 9^7$$ is divisible by 64.

Releted MCQ Question on
Algebra >> Binomial Theorem

Releted Question 1

Given positive integers $$r > 1, n > 2$$ and that the co - efficient of $${\left( {3r} \right)^{th}}\,{\text{and }}{\left( {r + 2} \right)^{th}}$$ terms in the binomial expansion of $${\left( {1 + x} \right)^{2n}}$$ are equal. Then

A. $$n = 2r$$

B. $$n = 2r + 1$$

C. $$n = 3r$$

D. none of these

View Answer

Releted Question 2

The co-efficient of $${x^4}$$ in $${\left( {\frac{x}{2} - \frac{3}{{{x^2}}}} \right)^{10}}$$ is

A. $$\frac{{405}}{{256}}$$

B. $$\frac{{504}}{{259}}$$

C. $$\frac{{450}}{{263}}$$

D. none of these

View Answer

Releted Question 3

The expression $${\left( {x + {{\left( {{x^3} - 1} \right)}^{\frac{1}{2}}}} \right)^5} + {\left( {x - {{\left( {{x^3} - 1} \right)}^{\frac{1}{2}}}} \right)^5}$$ is a polynomial of degree

A. 5

B. 6

C. 7

D. 8

View Answer

Releted Question 4

If in the expansion of $${\left( {1 + x} \right)^m}{\left( {1 - x} \right)^n},$$ the co-efficients of $$x$$ and $${x^2}$$ are $$3$$ and $$- 6\,$$ respectively, then $$m$$ is

A. 6

B. 9

C. 12

D. 24

View Answer

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

Releted MCQ Question on
Algebra >> Binomial Theorem

Given positive integers $$r > 1, n > 2$$ and that the co - efficient of $${\left( {3r} \right)^{th}}\,{\text{and }}{\left( {r + 2} \right)^{th}}$$ terms in the binomial expansion of $${\left( {1 + x} \right)^{2n}}$$ are equal. Then

The co-efficient of $${x^4}$$ in $${\left( {\frac{x}{2} - \frac{3}{{{x^2}}}} \right)^{10}}$$ is

The expression $${\left( {x + {{\left( {{x^3} - 1} \right)}^{\frac{1}{2}}}} \right)^5} + {\left( {x - {{\left( {{x^3} - 1} \right)}^{\frac{1}{2}}}} \right)^5}$$ is a polynomial of degree

If in the expansion of $${\left( {1 + x} \right)^m}{\left( {1 - x} \right)^n},$$ the co-efficients of $$x$$ and $${x^2}$$ are $$3$$ and $$- 6\,$$ respectively, then $$m$$ is

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Binomial Theorem

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

Releted MCQ Question on Algebra >> Binomial Theorem

Given positive integers $$r > 1, n > 2$$ and that the co - efficient of $${\left( {3r} \right)^{th}}\,{\text{and }}{\left( {r + 2} \right)^{th}}$$ terms in the binomial expansion of $${\left( {1 + x} \right)^{2n}}$$ are equal. Then

The co-efficient of $${x^4}$$ in $${\left( {\frac{x}{2} - \frac{3}{{{x^2}}}} \right)^{10}}$$ is

The expression $${\left( {x + {{\left( {{x^3} - 1} \right)}^{\frac{1}{2}}}} \right)^5} + {\left( {x - {{\left( {{x^3} - 1} \right)}^{\frac{1}{2}}}} \right)^5}$$ is a polynomial of degree

If in the expansion of $${\left( {1 + x} \right)^m}{\left( {1 - x} \right)^n},$$ the co-efficients of $$x$$ and $${x^2}$$ are $$3$$ and $$- 6\,$$ respectively, then $$m$$ is

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Algebra >> Binomial Theorem

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Binomial Theorem