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CAT 2023 Slot 3 — QA Question 5

FactorsEasy

The sum of the first two natural numbers, each having 15 factors (including 1 and the number itself), is

Answer & solution

Answer: 468

Solution

Easy

If a number's prime factorisation is p1e1p2e2⋯p_1^{e_1}p_2^{e_2}\cdots, its number of factors is (e1+1)(e2+1)⋯(e_1+1)(e_2+1)\cdots. We need this product to equal 15=1515=15 or 3×53\times5, which fixes the possible exponent patterns. Then assign the smallest primes to the largest exponents to get the two smallest such numbers.

1

Factor 15 to find the exponent patterns. Since the factor count must be 1515:

15=15 ⇒ a14or15=3×5 ⇒ a4⋅b215=15\ \Rightarrow\ a^{14}\qquad\text{or}\qquad 15=3\times5\ \Rightarrow\ a^{4}\cdot b^{2}
2

The a14a^{14} form is far too big:

214=163842^{14}=16384

so the smallest numbers come from the a4⋅b2a^4\cdot b^2 form.

3

Smallest two from a4b2a^4 b^2. Put the bigger exponent on the smaller prime:

24⋅32=16×9=14434⋅22=81×4=324\begin{aligned} &2^4\cdot 3^2=16\times 9=144\\ &3^4\cdot 2^2=81\times 4=324 \end{aligned}

(Any other prime pairing, e.g. 24⋅52=4002^4\cdot5^2=400, is larger, so these are the first two.)

4

Add them:

144+324=468144+324=468

Required sum =468=\mathbf{468}.

CAT 2023 Slot 3 QA Q5: The sum of the first two natural numbers, each having 15 factors (including 1 and the number itself), is — Solution | TheCATExam