Question

$3^{2\times 100+1}+2^{100+2}$ is divisible by________.

• A. 5
• B. 6
• C. 7
• D. 8

Answer 1

The correct option is C

7

When we are solving questions like this, we try to express the bases in terms of other numbers. In this case, the powers are written as 2×100+1 and 100+2. From this we can assume, we have to split the powers and proceed.

We will write $3^{2\times 100+1}$ as $3\times (3^2{)}^{100}$ and $2^{100+2}$ as $4\times 2^{100}$(We got this step after observing the way powers are given in question)

$\therefore$ $3^{2\times 100+1}$ + $2^{100+2}$ = $3\times (3^2{)}^{100}$ + $4\times 2^{100}$

= $3\times (9{)}^{100}$ + $4\times 2^{100}$

Now we have four numbers, 3, 4, 9 and 2. As a next step, we will try to express these numbers in some other common number like

9 = 6 + 3, 4 = 6 - 2, 3 =6 - 3 and 2 = 6 - 2. We can try many combinations like this. We have to decide in which way we should proceed.

We will proceed in the following way. It may not be clear in this step. If you can think two steps ahead/write the next two steps, you will understand the reason behind it.

$3\times (9{)}^{100}$ + $4\times 2^{100}$ = $3\times (7+2{)}^{100}$ + $(7-3)\times 2^{100}$

If you observe the last step, we can see that there is a chance of $3\times (2{)}^{100}$ getting cancelled.

$3\times (9{)}^{100}$ = $3\times (7+2{)}^{100}$ = 3$[{}^{100}{C}_0{7}^{100}+{}^{100}{C}_1{7}^{99}{2}^1+.........{}^{100}{C}_{99}{7}^1{2}^{99}+2^{100}]$

$\Rightarrow$ $3\times (7+2{)}^{100}$ + $(7-3)\times 2^{100}$ = 3$[{}^{100}{C}_0{7}^{100}+{}^{100}{C}_1{7}^{99}{2}^1+.........{}^{100}{C}_{99}{7}^1{2}^{99}+2^{100}]+7\times 2^{100}-3\times 2^{100}$

$\Rightarrow$ $3\times (7+2{)}^{100}$ + $(7-3)\times 2^{100}$ = 3$[multipleof7+2^{100}]+7\times 2^{100}-3\times 2^{100}$

$\Rightarrow$ $3\times (7+2{)}^{100}$ + $(7-3)\times 2^{100}$ = 3$\times multipleof7+3\times 2^{100}+7\times 2^{100}-3\times 2^{100}$

$\Rightarrow$ $3\times (7+2{)}^{100}$ + $(7-3)\times 2^{100}$ = 3$\times multipleof7+7\times 2^{100}+3\times 2^{100}-3\times 2^{100}$

$\Rightarrow$ $3\times (7+2{)}^{100}$ + $(7-3)\times 2^{100}$ = 3$\times multipleof7+7\times 2^{100}$

$\Rightarrow$ $3\times (7+2{)}^{100}$ + $(7-3)\times 2^{100}$ =$Multipleof7$

So it's divisible by 7.