The question where we had to find least positive integer having 24 factors?

Question

while solving case 3 why cant we take case as 2*3*4 which is 24 so the number would be 360 (2 with power 3 * 3 with power 2* 5 with power 1)
why only one case was taken which 2*2*6

Simran Mehta · Accepted Answer

Yes, you are correct. We can have the case with factors 2&times;3&times;4
Understanding the Relationship Between Prime Factorization and Number of Factors:If n has a prime factorization of the form n = p1^e1 * p2^e2 * ... * pk^ek, where p1, p2, ..., pk are distinct primes and e1, e2, ..., ek are their respective powers, then the total number of factors of n is given by:  (e1 + 1)(e2 + 1) ... (ek + 1)
Identify the Factor Combinations That Multiply to 24:- The number 24 can be expressed as a product of integers in several ways:  24 = 24 * 1  24 = 12 * 2  24 = 8 * 3  24 = 6 * 4  24 = 6 * 2 * 2  24 = 4 * 3 * 2
Choosing the Smallest Combination:To minimize n, we want to minimize the exponents and use the smallest primes possible.
Trying Different Combinations:- For 24 = 24 * 1:  - n = p1^23  - n = 2^23 (since 2 is the smallest prime)  - This gives a very large number.
- For 24 = 12 * 2:  - n = p1^11 * p2^1  - n = 2^11 * 3^1 = 2048 * 3 = 6144
- For 24 = 8 * 3:  - n = p1^7 * p2^2  - n = 2^7 * 3^2 = 128 * 9 = 1152
- For 24 = 6 * 4:  - n = p1^5 * p2^3  - n = 2^5 * 3^3 = 32 * 27 = 864
- For 24 = 6 * 2 * 2:  - n = p1^5 * p2^1 * p3^1  - n = 2^5 * 3^1 * 5^1 = 32 * 3 * 5 = 480
- For 24 = 4 * 3 * 2:  - n = p1^3 * p2^2 * p3^1  - n = 2^3 * 3^2 * 5^1 = 8 * 9 * 5 = 360
Identifying the Smallest Value:- By comparing the values from different combinations, the smallest positive integer with exactly 24 factors is 360.