If 10

Question

If  10 \leq x \leq 20  and  5 \leq y \leq 10 , and xy is a prime number, what is the least possible value of xy?

(A) 2
(B) 11
(C) 53
(D) 77
(E) Not possible

New Question

Tranquil Focus · Accepted Answer

chetan2u  , I am new to the forums. How can I retain format of algebraic formulae in my posts? Thanks. BTW- I've read a lot of your posts. You're an awesome teacher and your explanations have helped me learn a lot!

Since  10 <= x <= 20  and  5 <= y <= 10 ,

50<=xy<=200.

Now xy is a prime number- no way to get 2, no way to get 11 either. We can get 53 by taking y=5.3 and x=10. So least value of xy is 53. Hence C. Please confirm.

szeth · Answer

Interesting question. I don’t understand how xy can be a prime number as the very fact that the number is formed by multiplying x and y precludes it from being a prime number unless I’m reading the question wrongly.

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Sent from my iPhone using GMAT Club Forum

chetan2u · Answer

Hi..

It is nowhere mentioned x and y are integers..
I was about to add " not possible" as a choice, ill do it now

szeth · Answer

Ah fair enough.

@tranquil focus your method seems to make sense. I do not know how to give kudos via my phone but thanks for solving this

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Bunuel · Answer

Writing Mathematical Formulas on the Forum: https://gmatclub.com/forum/rules-for-po ... l#p1096628

exc4libur · Answer

A prime has 2 factors, {1,itself}. The stem doesn't say that x or y are integers.. thus, we can make a prime number multiplying with decimals.

The least possible value for xy = prime, has to be the first prime after the lower range  5*10=50 , which is  53=10*5.3  (10 is x and 5.3 is y).

(C) is the answer.

Kinshook · Answer

Given:  10 \leq x \leq 20  and  5 \leq y \leq 10 , and xy is a prime number.

Asked: What is the least possible value of xy?

50 \leq xy \leq 200

First prime number  \geq 50 = 53

IMO C

nirushanraj · Answer

Please correct me if I am wrong. If the question had stated that both x and y are integers, then 77 would be the correct answer, right?

chetan2u · Answer

No, that would not be correct. In that case xy will never be prime.

Adit_ · Answer

Nice question!
Most logical approach:
Prime number mentioned so hmm no co prime, no cancellations with respective to the rational number itself.
So it means the number stands tall in numerator with nothing to bring it down!
Now we also know lowest possible product here is 50.

Using these info we can check if 53 works as its in options:
53/x*x is the form.
Take lowest value which is 10 so 53/10=5.3 (falls within range of >=5)
10 gets cancelled out thus 53 is valid indeed.

Answer: Option C

egmat · Answer

Hi,

This problem has a classic  trap  — let me walk you through it.

We know  10  ≤ x ≤  20  and  5  ≤ y ≤  10 , and xy must be a  prime number .

Step 1:  Find the range of xy.
The smallest possible product is  10  ×  5  =  50 .
The largest possible product is  20  ×  10  =  200 .
So xy must be somewhere between  50  and  200 .

Step 2:  Here's the   KEY insight   — the problem never says x and y are integers! They can be any real numbers within their ranges (like  10.5 ,  7.3 , etc.).

Step 3:  Since xy can be ANY value from  50  to  200  (not just integer products), we just need the smallest  prime number  that is ≥  50 .

Primes near  50 :  47  (too small),  53  ✓

So  53  is our answer.

Step 4:  Verify — can we actually make xy =  53 ?  Yes! Set x =  10  and y =  5.3 . Both are within their allowed ranges, and  10  ×  5.3  =  53 , which is prime. ✓

Answer: C (53)

The Trap:  If you assume x and y must be integers, then any product of two integers where both are greater than  1  can never be prime (since primes have no factors other than  1  and themselves). That would lead you to pick E (Not possible) — the most common wrong answer.

General Principle:  On GMAT, if a problem doesn't explicitly state that variables are integers (or positive integers), they can be decimals, fractions, or negative numbers. Always check what constraints are actually given!

If 10 <= x <= 20 and 5 <= y <= 10, and xy is a prime number, which is

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