Question

If $P=21\left( {{21}^{2}}-{{1}^{2}} \right)\left( {{21}^{2}}-{{2}^{2}} \right)\left( {{21}^{2}}-{{3}^{2}} \right)...\left( {{21}^{2}}-{{10}^{2}} \right)$, the P is divisible by

Answer 1

In this question, we are asked to find out the P divisibility, and we are also learnt about definition of divisibility rules, and we reduce the given expression$P=21\left( {{21}^{2}}-{{1}^{2}} \right)\left( {{21}^{2}}-{{2}^{2}} \right)\left( {{21}^{2}}-{{3}^{2}} \right)...\left( {{21}^{2}}-{{10}^{2}} \right)$to find out the P divisibility.

Complete step-by-step solution:
Divisibility rules are the rules which help in determining whether a given number is divisible by the given divisor or not, without actually performing the division.
There are many divisibility rules like divisibility by 2, divisibility by 3, divisibility by 4, divisibility by 5, divisibility of 7, divisibility of 9 and divisibility of 11 and so on.
Let us solve our given expression to find out P divisibility
Given expression,
$P=21\left( {{21}^{2}}-{{1}^{2}} \right)\left( {{21}^{2}}-{{2}^{2}} \right)\left( {{21}^{2}}-{{3}^{2}} \right)...\left( {{21}^{2}}-{{10}^{2}} \right)$
Above expression is expanded by using the formula $\left( {{a}^{2}}-{{b}^{2}} \right)=\left( a+b \right)\left( a-b \right)$, we get
$\Rightarrow P=21\left( 21+1 \right)\left( 21-1 \right)\left( 21+2 \right)\left( 21-2 \right)...\left( 21+10 \right)\left( 21-10 \right)$
by re-arranging, we get
$\Rightarrow \left( 21-10 \right)\left( 21-9 \right)\left( 21-8 \right)........\left( 21-0 \right)\left( 21+1 \right)........\left( 21+10 \right)$
So, we subtract the individuals we get
$\Rightarrow 11.12.13.14............30.31$
Hence, the total terms are 21 consecutive terms
The above terms are in the manner of
$\Rightarrow n.\left( n+1 \right)\left( n+2 \right)\left( n+3 \right)...........$
Or else, we can also write this property
$P=r\left( r+1 \right)\left( r+2 \right)\left( r+3 \right)...........\left( r+n-1 \right)$
The number of ways is in the form of $P=r\left( r+1 \right)\left( r+2 \right)\left( r+3 \right)...........\left( r+n-1 \right)$
${}^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}$
${}^{n}{{P}_{r}}={}^{n}{{C}_{r}}\times r!$
$where\text{ }{}^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!r!}$
Or we can write as $\text{P}=\dfrac{n!}{\left( n-r \right)!r!}\cdot r!$
If we substitute the n and r values, we can get the solution,
Here, $\begin{aligned} & n=31 \\\ & r=21 \\\ \end{aligned}$
Substituting the above terms, we get

& P=\dfrac{31!}{10!21!}21! \\\ & P=21!{}^{31}{{C}_{10}} \\\ \end{aligned}$$ Hence, P is divisible by $$21!$$. **We can conclude that P is divisible by $$21!$$.** **Note:** We need to know the permutation and combinations formulas, so that we can easily solve these kinds of problems. If you a big divisor, break into its prime factor and check the divisibility. If the dividend is divisible by all the prime factors of the divisor, then the number is divisible.