Increase in number of factors

In order to solve this question, we must understand why the increase in factors is happening, i.e. how it is caused.

For purposes of illustration, let us consider the number 33 .

33 has 4 factors-1,3,11,33

33*2=66 has 8 factors-1,2,3,6,11,22,33,66. The 'new' factors are 2,6,22,66 .

We observe that all the new factors can be expressed as (Any old factor*2) . Note that factors formed by this method do not necessarily have to be 'new'. For eg. 4 has factors-1,2,4 while 8 has factors 1,2,4,8 with only 8 being a new factor as factors1,2 on multiplication by 2 did not yield new factors.

In this problem, we assume that among the 15 factors for the original number n, there are only 5 such numbers which on multiplication by 2 will yield new factors. Let us write the 15 factors as w1,w2,w3....w10 f1,f2...,f5 where factors with the letter 'f' on multiplication by 2 will yield new factors.

Thus,

n has factors w1,w2,....w10,f1,f2...f5 15 factors

2n has factors w1,w2....w10,f1,f2...f5,2(f1),2(f2)....2(f5) 15+5 factors

4n has factors w1,w2....w10,f1,f2...f5,2(f1),2(f2)....2(f5)...4(f1),4(f2)...4(f5) 20+5 factors

Thus the answer is 25.

If n = 15 factors, 2n = 20 factors (15 +5), then 4n = 15 +5x2 as 4n is a double of 2n

Considering the amount of divisors $n$ has (including 1, and n itself), we deduce n must be of one of the following forms:

• 1) $N = p_1^{14}$
• 2) $N = p_1^{4} \times p_2^{2}$

Because only then, we have 15 different different integers, which obviously each of them divide n. (We include 0 too as a possiblity, hence 1 is a divisor too) If we repeat the process for a number to have 20 factors, we will conclude there are three possiblities, though only one of them is possible considering our previous assumptions, which is $2N = p_1^{3} \times p_2^{4}$ .

Again it shows that there are 20 divisors using rule of product, and more over it is a possible consequence of case (2), assuming $p_2 = 2, 2^{2} \times 2 = 2^{3}$ . Observing $4N = 2 \times 2N$ . We will multiply $p_2$ by 2 again, and notice it's exponent is 4 now. Considering all the possibilities, there are: $5 \times 5 = 25$ different divisors.

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