GATE CSE 2015 Set 2 | Question: 9

Question

The number of divisors of $2100$ is ____.

Comments

https://gateoverflow.in/2015/gate2014-2-49

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HER THIS MIGHT HELP ARE REALLY HELPFUL THANKS

Answer 1

Answer: 36

$2100 = 7\times 3\times 2^2 \times 5^2$

Hence, total number of factors will be $= (1+1)\times (1+1)\times (2+1)\times (2+1) \\= 2 \times 2\times 3 \times 3 \\= 36$,

because any factor is obtained by multiplying the prime factors zero or more times. (one extra for zero)

Comments

D = qd + 0;

In this familiar equation "d" is the divisor,

if q = [-2100] and d = [-1]  then,  D = 2100.   So [-1] and [-2100] can also be divisors.

So total = 72.

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http://mathschallenge.net/library/number/number_of_divisors

is a good read

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          2100
           /   \
        2    1050
                /  \
             2    525
                    /  \
                  3    175
                         /  \
                        5   35
                              / \
                            5   7
                      
2² 3¹ 5²  7¹   => (2+1)*(1+1)(2+1)*(1+1) => 3*2*3*2 = 36
 

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@Dq Our objective Should here to do prime factorization...can be done in many ways

2100=21*100=3*7*$5^2$*$2^2$

Answer 2

 

$2100 = 7 * 3 * 2^{2} * 5^{2}$

To calculate the total number of divisors, take each power in the prime factorization obtained, and:

Step 1: Increment the power.

$(1+1) (1+1) (2+1) (2+1)$

Step 2: Calculate the product of it:

$(1+1) * (1+1) * (2+1)* (2+1)$

Answer: 36