Question

Find out by which number ${{7}^{9}}+{{9}^{7}}$ is divisible by
A. 128
B. 24
C. 64
D. 72

Answer 1

Hint: First take ${{7}^{9}}$ and ${{9}^{7}}$ separately then break them as ${{\left( 8-1 \right)}^{9}}$ and ${{\left( 8+1 \right)}^{7}}$ respectively using binomial theorem which states that,
${{\left( a+b \right)}^{n}}=C_{0}^{n}{{a}^{n}}+C_{1}^{n}{{a}^{n-1}}b+C_{2}^{n}{{a}^{n-2}}{{b}^{2}}+...........+C_{n}^{n}{{b}^{n}}$
Then find the remainder by dividing 64 and then add up to get solution

Complete step-by-step answer:
In the question we have to find the division of ${{7}^{9}}+{{9}^{7}}$_.
So, let ${{7}^{9}}+{{9}^{7}}$ be as (a + b) where $a={{7}^{9}}$ and $b={{9}^{7}}$_.
Now we will first operate on ‘a’ which is ${{7}^{9}}$ so we can represent it as,
$a={{\left( 8-1 \right)}^{9}}$
Now we will use the binomial theorem which states as,
${{\left( a-b \right)}^{n}}=C_{0}^{n}{{a}^{n}}-C_{1}^{n}{{a}^{n-1}}b+C_{2}^{n}{{a}^{n-2}}{{b}^{2}}-C_{3}^{n}{{a}^{n-3}}{{b}^{3}}+...........+C_{n}^{n}{{b}^{n}}$
Now using the formula we get,
${{\left( 8-1 \right)}^{9}}=C_{0}^{9}{{8}^{9}}-C_{1}^{9}{{8}^{8}}+C_{2}^{9}{{8}^{7}}-C_{3}^{9}{{8}^{6}}+.............C_{7}^{-9}{{8}^{2}}+C_{8}^{9}8-C_{9}^{9}1$
Now by analyzing all the terms separately we can say that when a is divided by 64 it divides all the terms separately except the last two.
Now by adding the last two terms we get,
$9\times 8-1=71$
If 71 is divided by 64 it gives 7 as remainder.
Now we write the given information in the form a = b (mod c) which represents that when a is divided by C it gives b as remainder.
So, we can write ${{7}^{9}}$ as,
${{7}^{9}}=7\left( \bmod 64 \right)$………………. (i)
Now let's take b to operate which can represent as,
${{9}^{7}}={{\left( 8+1 \right)}^{7}}$
Now we will use the binomial theorem which states that,
${{\left( a+b \right)}^{n}}=C_{0}^{n}{{a}^{n}}+C_{1}^{n}{{a}^{n-1}}b+C_{2}^{n}{{a}^{n-2}}{{b}^{2}}+..........+C_{n}^{n}{{b}^{n}}$
Now using the formula we get,
${{\left( 8+1 \right)}^{7}}=C_{0}^{7}{{8}^{7}}+C_{1}^{7}{{8}^{6}}+C_{2}^{7}{{8}^{5}}+.........+C_{6}^{7}8+C_{1}^{7}1$
Now analyzing each term separately we can say that except the last two terms all the other terms are divisible by 64.
So the sum of last number represents the remainder which we get,
$7\times 8+1=57$
Here 57 is the remainder.
So we can write that,
${{9}^{7}}=57\left( \bmod 64 \right)$ ……………. (ii)
Now from equation (i) and (ii) we can write that,
${{7}^{9}}=64k+7$ where K is a finite integer.
${{9}^{7}}=64l+57$ where l is also a finite integer.
Now we can add ${{7}^{9}}$ and ${{9}^{7}}$in terms of k and l so we get,
${{7}^{9}}+{{9}^{7}}=64k+7+64l+57$
We can say that,
${{7}^{9}}+{{9}^{7}}=64\left( k+l+1 \right)$
 As we know k, l is finite number so (k + l + 1)
As we know k, l is a finite number.
${{7}^{9}}+{{9}^{7}}=64\left( k+l+1 \right)$
As we know k, l is a finite number so (k + l +1) is also a finite number.
Hence the correct option is ‘C’.

Note: Students should know about the binomial theorem before doing this. They should also be careful above the calculations to avoid any errors or mistakes.