Question

The remainder when ${1690}^{2608}+{2608}^{1609}$ is divided by 7, is

• A. 2
• B. 3
• C. 1
• D. None of these

Answer 1

The correct option is C

1

In this case, we have to express 1690 and 2608 in terms of multiples of 7.

1690 = 7 $\times$ 241 + 3 and 2608 = 7 $\times$ 372 + 4

$\Rightarrow$ ${1690}^{2608}+{2608}^{1609}$ = $(7\times 241+3{)}^{2608}+(7\times 372+4{)}^{1609}$

When we expand $(7\times 241+3{)}^{2608}$, except the last term, every other term will be a multiple of 7.So we can write it as 7k + $3^{2608}$

(The last term is $3^{2608}$)

Similarly,we can write $(7\times 372+4{)}^{1609}$ as 7m + $4^{1609}$

Now the sum ${1690}^{2608}+{2608}^{1609}=7k+3^{2608}+7m+4^{1609}$

=$7(k+m)+3^{2608}+4^{1609}$

To find the remainder, we will consider the terms $3^{2608}+4^{1609}$ only.We have to express both the terms in terms of multiples of 7.

We know $3^3$ = 27 = 28 - 1 = 4 * 7 -1

Also, $4^3=64=63+1=9^{\ast }7+1$

$3^{2608}+4^{1609}=3\times 3^{3\times 867}+4\times 4^{3\times 563}$

$3^{2608}+4^{1609}=3\times {27}^{867}+4\times {64}^{563}$

$3^{2608}+4^{1609}=3\times (28-1{)}^{867}+4\times (63+1{)}^{563}$

In the expansion of $(28-1{)}^{867}$,except the last term,every term will be a multiple of 28(and 7).So we can write $(28-1{)}^{867}$ as

$(28-1{)}^{867}$ = 7p - 1

Similarly, $(63+1{)}^{563}$ can also be written as

$(63+1{)}^{563}$ = 7q + 1

$\Rightarrow$ $3^{2608}+4^{1609}=3\times (7p-1)+4\times (7q+1)$

= 7(3p + 4q) + 1

$\Rightarrow$ The remainder is 1