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02 Jul 2025, 21:05
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The total # of factors can be obtain by expressing the number in the prime factorization form and then multiplying 1+ the exponent of each prime factor.
example
12 = 2^2 * 3
Total factors: (2+1) (1+1) = 6
For max the number of prime factors i must express 24 in the longest multiplication form : 2 * 2 * 2 * 3
That is the same as (1+1)(1+1)(1+1)(2+1),
each parenthesis means one prime factor, so 4 is the max number of prime factors
But the question ask the range, so i need to also find the min number of prime factors.... simple.... (23+1) equals 24 and means 1 prime factor.
So the range is 4-1 = 3
example
12 = 2^2 * 3
Total factors: (2+1) (1+1) = 6
For max the number of prime factors i must express 24 in the longest multiplication form : 2 * 2 * 2 * 3
That is the same as (1+1)(1+1)(1+1)(2+1),
each parenthesis means one prime factor, so 4 is the max number of prime factors
But the question ask the range, so i need to also find the min number of prime factors.... simple.... (23+1) equals 24 and means 1 prime factor.
So the range is 4-1 = 3
02 Jul 2025, 21:05
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Bunuel
We know that the number of positive factors of a number M whose prime factorised form is \({P_1}^x* {P_2}^y*........ {P_n}^z\) is \((x+1)(y+1).....(z+1)\)
so the minimum number of prime factor could be 1, as the number could be of the form \({P_1}^{23}\)
To maximize the number of distinct prime factors, we must factor 24 as a product of as many small integers greater than 1 as possible, since each factor corresponds to an exponent plus 1. What we are basicall doing is maximizing the number of terms in:
\((x+1)(y+1).....(z+1)\)
so, 24= \(2^3*3^1\) , thus 3+1=4 maximum prime factors
now we are asked the range, which would be 4-1=3
Thus, IMO C.
02 Jul 2025, 21:13
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To find no of prime factors we do prime factorization so in this case it would be 24=3*2*2*2. Now we need to find the range of maximum possible in that we subtract min and max no of prime factors. So minimum is always 1 ie. the number itself. Range would be 4-1=3 Option C
Bunuel
