Question

Find the smallest positive integer such that it has exactly 100 different positive integer divisors including

1 and the number itself

Answer 1

we know That total No. of factors
=product of (prime no's power+1)

100 = 2²
× 5²= 2 × 2 × 5 × 5 = (1 + 1) (1 + 1) (4 + 1) (4 + 1)
Let the number be N
then N=p¹×q¹×r⁴×s⁴
=5×7×2⁴×3⁴
=45360
hence the smallest number is 45360
Hope this helps you...

Answer 2

Answer:

Every composite number can be expressed as a product of prime factors.

If N is the number of different divisors:

N = ( k1 + 1 ) · ( k2 + 1 ) ··· ( kn + 1 )

Then the integer n = p1^k1 · p2^k2 ···· pn^kn

So : N = 100

100 = 4 · 25 = = 2 · 2 · 5 · 5 = ( 1 + 1 ) · ( 1 + 1 ) · ( 4 + 1 ) · ( 4 + 1 )

For the smallest value: k1 = 4, k2 = 4, k3 = 1, k4 = 1.

Finally:   n = p1^4 · p2^4 · p3^4 · p4^4 = 2^4 · 3^4 · 5^1 · 7^1 =

= 16 · 81 · 5 · 7 = 45,360

Answer: The number is 45,360.

Step-by-step explanation:

hope this helps u

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