Question

Find the number of divisors and sum of divisors of number 1400. Also find the number of ways of putting 1400 as a product of two factors.

Answer 1

Hint: To solve this question, we will represent 1400 as powers of prime number and then we will apply the formulas for the number of divisors, the sum of divisors and formula for resolving the number as a product of two factors.

Complete step-by-step answer:
Before solving the question, we must know what is a divisor of a number. Divisor of a number $'n'$ is the number which leaves no remainder when ‘n’ is divided by that number. In this question, we have three parts to calculate. Before starting to calculate them, we will first represent 1400 in terms of power of prime numbers. Thus, 1400 can also be written as
$\begin{align}
  & 2\left| \!{\underline {\,
  1400 \,}} \right. \\
 & 2\left| \!{\underline {\,
  700 \,}} \right. \\
 & 2\left| \!{\underline {\,
  350 \,}} \right. \\
 & 5\left| \!{\underline {\,
  175 \,}} \right. \\
 & 5\left| \!{\underline {\,
  35 \,}} \right. \\
 & 7\left| \!{\underline {\,
  7 \,}} \right. \\
 & \text{ 1} \\
\end{align}$
Hence we can say that, $1400=2\times 2\times 2\times 5\times 5\times 7$
$\Rightarrow 1400={{2}^{3}}\times {{5}^{2}}\times 7$
Now, we will calculate the number of divisor. The formula for calculating the total number of divisor of a number $'n'$ where n can be represent as powers of prime numbers is shown as.
If $N={{p}^{a}}{{q}^{b}}{{r}^{c}}$ .
Then total number of divisors \[=\left( a+1 \right)\left( b+1 \right)\left( c+1 \right).\] In our case, \[a=3,\text{ }b=2\text{ }and\text{ }c=1\] . Thus total number of divisors
$=\left( 3+1 \right)\left( 2+1 \right)\left( 1+1 \right)$
$=4\times 3\times 2$
$=24$ .
Now, we will calculate the sum of divisors of 1400. The formula for finding the sum of divisors is given as:
Sum of divisors $=\left( {{P}^{0}}+{{P}^{1}}+{{P}^{2}}......{{P}^{a}} \right)\left( {{q}^{0}}+{{q}^{1}}+{{q}^{2}}......{{q}^{b}} \right)\left( {{r}^{0}}+{{r}^{1}}+{{r}^{2}}......{{r}^{c}} \right)$ .
In our case, the sum of divisors,
Sum of divisors $=\left( {{2}^{0}}+{{2}^{1}}+{{2}^{2}}+{{2}^{3}} \right)\left( {{5}^{0}}+{{5}^{1}}+{{5}^{2}} \right)\left( {{7}^{0}}+{{7}^{1}} \right)$
$=\left( 1+2+4+8 \right)\left( 1+5+25 \right)\left( 1+7 \right)$
$=15\times 31\times 8$
$=3720.$
Now, we will see how we can resolve 1400 as a product of two factors. The formula for resolving a number (which is not a square of an integer) into product of two factors is $=\dfrac{1}{2}\left[ \left( a+1 \right)\left( b+1 \right)\left( c+1 \right) \right]$ .
Thus is our case, the value will be $=\dfrac{1}{2}\left[ \left( 3+1 \right)\left( 2+1 \right)\left( 1+1 \right) \right]$
$=12.$
Hence, total number of divisors = 24.
Sum of divisors = 3720.
Number of ways of putting 1400 as a product of two factors =12.

Note: In the calculation of the third part, the formula is valid only for those number which are not squares of integers if we were given a number which is square of an integer then the formula $=\dfrac{1}{2}\left[ \left( a+1 \right)\left( b+1 \right)\left( c+1 \right)+1 \right]$ .