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CAT 2018 Slot 2 — QA Question 33

Remainders BasicsEasy

If A = {62n - 35n - 1: n = 1, 2, 3, ...} and B = {35(n - 1) : n = 1,2,3,...} then which of the following is true?

Answer & solution

  • Every member of A is in B and at least one member of B is not in A

  • B

    At least one member of A is not in B

  • C

    Every member of B is in A.

  • D

    Neither every member of A is in B nor every member of B is in A

Solution

Easy

Show every element of AA is a multiple of 3535 (so it lies in BB, the set of all non-negative multiples of 3535), then exhibit a multiple of 3535 that AA misses. That gives A⊆BA\subseteq B with the containment strict.

1

Describe BB. B={35(n−1):n=1,2,3,… }={0,35,70,105,… }B=\{35(n-1):n=1,2,3,\dots\}=\{0,35,70,105,\dots\} — every non-negative multiple of 3535.

B={0, 35, 70, 105, … }\begin{aligned} &B = \{0,\ 35,\ 70,\ 105,\ \dots\} \end{aligned}
2

Every element of AA is a multiple of 3535. Rewrite 62n=36n6^{2n}=36^n.

62n−35n−1=(36n−1)−35n⇒ 36n−1 is divisible by 36−1=35(an−1 divisible by a−1)⇒ 35n is divisible by 35⇒ every element of A is a multiple of 35, so A⊆B\begin{aligned} &6^{2n} - 35n - 1 = (36^n - 1) - 35n\\ &\Rightarrow\ 36^n - 1\ \text{is divisible by } 36-1 = 35 \quad\text{(}a^n-1\text{ divisible by }a-1\text{)}\\ &\Rightarrow\ 35n\ \text{is divisible by } 35\\ &\Rightarrow\ \text{every element of }A\ \text{is a multiple of } 35,\ \text{so } A\subseteq B \end{aligned}
3

Containment is strict. Compute the first few elements of AA and note a gap.

n=1: 36−35−1=0n=2: 1296−70−1=1225=35⋅35⇒ 35∈B but 35∉A(A jumps 0→1225)\begin{aligned} &n=1:\ 36 - 35 - 1 = 0\\ &n=2:\ 1296 - 70 - 1 = 1225 = 35\cdot 35\\ &\Rightarrow\ 35\in B\ \text{but } 35\notin A \quad\text{(A jumps }0\to 1225\text{)} \end{aligned}

Every member of AA is in BB, and at least one member of BB (e.g. 3535) is not in AA.

A⊊BA \subsetneq B
CAT 2018 Slot 2 QA Q33: If A = {6 2n - 35n - 1: n = 1, 2, 3, ...} and B = {35(n - 1) : n = 1,2,3,...} then which of the following is t — Solution | TheCATExam