TheCATExam
Sign in

CAT 2023 Slot 1 — QA Question 14

FactorsEasy

Let n be the least positive integer such that 168 is a factor of 1134n. If m is the least positive integer such that 1134n is a factor of 168m, then m + n equals

Answer & solution

Correct answer: 15

  • A

    24

  • B

    12

  • 15

  • D

    9

Solution

Easy

Divisibility is a prime-by-prime game. Factorise both bases, then for each direction compare exponents prime-by-prime: the tightest constraint fixes the least nn, then the least mm.

1

Prime-factorise:

168=23⋅3⋅71134=2⋅34⋅7\begin{aligned} &168=2^3\cdot 3\cdot 7\\ &1134=2\cdot 3^4\cdot 7 \end{aligned}
2

Least nn with 168∣1134 n168 \mid 1134^{\,n}. Here 1134 n=2 n⋅34n⋅7 n1134^{\,n}=2^{\,n}\cdot 3^{4n}\cdot 7^{\,n} must contain 23⋅31⋅712^3\cdot 3^1\cdot 7^1:

2: n≥3,3: 4n≥1,7: n≥1⇒ n=3  (the 2-power binds)\begin{aligned} &2:\ n\ge 3,\qquad 3:\ 4n\ge 1,\qquad 7:\ n\ge 1\\ &\Rightarrow\ n=3\ \ (\text{the }2\text{-power binds}) \end{aligned}
3

Least mm with 1134 n∣168 m1134^{\,n}\mid 168^{\,m}. Now 11343=23⋅312⋅731134^{3}=2^{3}\cdot 3^{12}\cdot 7^{3} must divide 168 m=23m⋅3m⋅7m168^{\,m}=2^{3m}\cdot 3^{m}\cdot 7^{m}:

2: 3m≥3,3: m≥12,7: m≥3⇒ m=12  (the 3-power binds)\begin{aligned} &2:\ 3m\ge 3,\quad 3:\ m\ge 12,\quad 7:\ m\ge 3\\ &\Rightarrow\ m=12\ \ (\text{the }3\text{-power binds}) \end{aligned}
4

Add:

m+n=12+3=15m+n=12+3=15

m+n=15m+n=15 — option (c).

Related Factors questions

See all Number Theory questions →
CAT 2023 Slot 1 QA Q14 — Factors solution | TheCATExam