If S is a set of odd integers and 3 and -1 are in S, is -15 in S? (1) 5 is in S. (2) Whenever two numbers are in S, their product is in S. DS10602.01

Answer Choices

Correct answer marked below

Statement (1) ALONE is sufficient

Statement (2) ALONE is sufficient

BOTH statements TOGETHER are sufficient

Correct

EACH statement ALONE is sufficient

Statements (1) and (2) TOGETHER are NOT sufficient

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Comments

SushilJun 27

Thanks! I was overcomplicating this. I initially thought statement (2) alone might be enough but then realized we need an actual number to start with. The combination of 3, -1, and 5 does the trick.

ZofishanJul 5

Same here! I was trying to see if -15 could be formed from just 3 and -1 using the product rule, but that only gives 3, -1, -3, etc. Adding 5 makes the difference.

AhmedJun 29

Exactly. The key is that statement (2) alone doesn't give any new numbers because we don't know what's in S beyond 3 and -1. We need statement (1) to bring in 5.

ZachariahJun 3

Why isn't statement (2) alone sufficient? If 3 and -1 are in S, then their product -3 is in S, and then 3 times -3 is -9, and -1 times -3 is 3, etc. But we never get -15. So (2) alone doesn't help. But with (1) giving 5, we can get -15 via 3 * 5 = 15 and then 15 * -1 = -15. That works because -1 is in S. So both together are sufficient. Answer is C.

KaushikJun 12

Nice breakdown. I missed that we could use 5 and -1 to get -5, then 3 and -5 to get -15. But your way with 15 and -1 is simpler.

NickMay 12

This one was a bit tricky because I had to think about closure under multiplication. At first I thought statement (2) alone might be enough if we could generate 5 from 3 and -1, but we can't. So we definitely need statement (1). Good question.

What this question tests

What it tests

Your ability to spot and extend arithmetic and geometric sequences and work with recursive definitions.

Common trap

Off-by-one errors in indexing (a₀ vs a₁) or confusing an arithmetic difference with a geometric ratio.