How many divides me?

Number Theory Level 2

Find the number of distinct positive divisors of 21600 (inclusive of 1 and itself).

72 60 75 45

1 solution

Abid Junaid
May 23, 2015

To find the total no of divisors of N.

write the prime factors on N

if N = a^p .b^y. c^r and so on then total no of divisors of

N = (a+1)(b+1)(c+1)

21600 = 2^5 X 3^3 X 5^2 (^ denotes power)

so total no of divisors = (5+1)(3+1)(2+1) = 6 X 4 X 3 = 72

Moderator note:

Please check your working again.

Do you mean "The number of divisors of $N$ is (p+1)(y+1)(r+1)"?

How are you able to factor the number so quickly?

Why must the answer be "(p+1)(y+1)(r+1)"?

Here's how I did it:

$21600 = 216 \times 100 = 6^3 \times 10^2 = (2 \times 3)^3 \times (2\times 5)^2$ .

Equivalently, $21600 = 2^{3+2} \times 3^3 \times 5^2 = 2^5 \times 3^3 \times 5^2$ .

The number of factors (inclusive of 1 and itself) of the $N = p_1^{r_1} \times p_2^{r_2} \times p_3^{r_3}$ is simply equals to $(r_1 + 1)(r_2+ 1)(r_3+1)$ .

In this case $p_1 = 2, p_2 = 3, p_3 = 5, r_1 = 5, r_2 = 3, r_3 = 2$ .

Hence the answer is $(5+1)(3+1)(2+1) = 72$ factors.

How many divides me?

1 solution

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