If an and b are chosen at random from the set of integers from 2 to 9

Question

If  a  and  b  are each chosen at random from the set of integers from 2 to 9, inclusive, what is the probability that  \frac{96a}{b}  is the square of an integer?

(A)  \frac{1}{16} 
(B)  \frac{1}{8} 
(C)  \frac{3}{16} 
(D)  \frac{1}{4} 
(E)  \frac{3}{8}

raffaeleprio · Answer

Let's rewrite the problem as:

96*\frac{a}{b} = n^2    =>   2^4*2*3* \frac{a}{b} = n^2

We can directly exclude the primes 5 and 7 for  a  or  b  since we can play only with the integers whose factors are 2 or 3.

Here we have limited possibilities to make the LHS a perfect square:

a=3  and  b=2  and viceversa (2 options)
 a=3  and  b=8  and viceversa (2 options)
 a=4  and  b=6  and viceversa (2 options)
 a=9  and  b=6  (1 option)

Total  7  options make the equation possible.

In addition, the number of possibilities for the draw of couples (order important) between 2 and 9 inclusive is  2!*\binom{8}{2} = 56

Hence probability  \frac{1}{8}

IMO A!

kevincan · Answer

Fantastic effort, you only overlooked 6,9 and assumed that an and b had to be distinct. 
I have tweaked the wording and the answer choices

Edskore · Answer

The key concept here is perfect squares via prime factorisation — recognising that 96a/b is a perfect square only when every prime in the result has an even exponent.

Key setup: 96 = 2^5 x 3. For 96a/b to be a perfect square integer, b must divide 96a, and all prime exponents in the result must be even. Since a and b come from {2,3,4,5,6,7,8,9}, any 5 or 7 in either variable cannot be cancelled — so no valid pair involves 5 or 7.

Step 1 — Count total outcomes

a and b are chosen independently from {2,3,4,5,6,7,8,9}, so total outcomes = 8 x 8 = 64. Note: order matters — (3,2) and (2,3) are different draws, and a = b is allowed.

Step 2 — Enumerate valid (a, b) pairs

(3, 2): 96 x 3/2 = 144 = 12^2 ✓
(2, 3): 96 x 2/3 = 64 = 8^2 ✓
(6, 4): 96 x 6/4 = 144 = 12^2 ✓
(4, 6): 96 x 4/6 = 64 = 8^2 ✓
(9, 6): 96 x 9/6 = 144 = 12^2 ✓
(6, 9): 96 x 6/9 = 64 = 8^2 ✓
(3, 8): 96 x 3/8 = 36 = 6^2 ✓
(8, 3): 96 x 8/3 = 256 = 16^2 ✓

Total: 8 valid pairs.

Step 3 — Calculate probability

P = 8/64 = 1/8

Answer: B

---

Common trap: raffaeleprio's approach above is excellent but treats a and b as distinct (without replacement), giving 56 total arrangements instead of 64. Since the problem says "each chosen at random" independently, repetition is allowed — total sample space is 8 x 8 = 64, not C(8,2) x 2. Also easy to overlook (8,3): 96 x 8/3 = 256 = 16^2.

Takeaway: On prime factorisation probability problems, enumerate systematically and always check whether the draws are with or without replacement before setting up your denominator.

(Kavya | 725 on GMAT Focus Edition)

dec2023 · Answer

96a/b needs to be perfect square

96a/b = 3x(2^5)x(a/b)

for 96a/b to be perfect square, all exponents must be even. So, (a/b) needs to multiple of 3x2=6.

within set 2-9, only 1 multiple of 6 =>6
total numbers in set = 8
total number of multiple of 6 = 1
so, probability = 1/8

GraemeGmatPanda · Answer

We first calculate the total number of possible ordered pairs  (a,b)  since each choice is independent and each can be any of the 8 integers from 2 to 9.
 8 	imes 8 = 64

We find the prime factorization of 96 to prepare for applying the exponent conditions.
 96 = 2 	imes 48 
 48 = 2 	imes 24 
 24 = 2 	imes 12 
 12 = 2 	imes 6 
 6 = 2 	imes 3 
 = 2^5 	imes 3

We express  96^{\frac{a}{b}}  in terms of its prime factors raised to fractional exponents.
 96^{\frac{a}{b}} = (2^5 	imes 3)^{\frac{a}{b}} 
 = 2^{5\frac{a}{b}} 	imes 3^{\frac{a}{b}}

For  96^{\frac{a}{b}}  to be a perfect square, the exponents of both primes in its prime factorization must be even integers.
 3^{\frac{a}{b}} 	ext{ implies }\frac{a}{b} 	ext{ is an integer} 
 2^{5\frac{a}{b}} 	ext{ implies }5\frac{a}{b} 	ext{ is even}\Rightarrow \frac{a}{b} 	ext{ is even}

Let  k = \frac{a}{b}  be an even integer. We count how many pairs  (a,b)  satisfy this within the given range.
 k = \frac{a}{b},\, k 	ext{ even} 
 k = 2: 
 a = 2b \le 9\Rightarrow b \le 4.5\Rightarrow b = 2,3,4\;(3	ext{ outcomes}) 
 k = 4: 
 a = 4b \le 9\Rightarrow b \le 2.25\Rightarrow b = 2\;(1	ext{ outcome}) 
 	ext{Total favorable} = 3 + 1 = 4

We then apply the probability formula to find the probability of these favorable outcomes out of the total.
 P = \frac{4}{64} = \frac{1}{16}

Answer A

Hope this helps! 
ʕ•ᴥ•ʔ

kevincan · Answer

I think the previous poster misinterpreted 96a divided by b as 96 to the power of a/b

HarshavardhanR · Answer

https://gmatclub.com/forum/download/file.php?id=87936 --- Harsha GMAT-Club-Forum-s7fg5z3u.png

kevincan · Answer

Transcribing Harsha :

(a, b) are chosen from 2 to 9 inclusive (8 possibilities each)

Total outcomes:
T = 8 × 8

P = F / T

96a/b is a perfect square

96 = 2^5 × 3

So 96a/b = 2^5 × 3 × a/b

A perfect square means even powers

Cases:

a b
2 3
3 2
3 8
4 6
6 4
6 9
8 3
9 3

8 cases

F = 8

P = F / T = 8 / (8 × 8) = 1/8

Kinshook · Answer

If and b are each chosen at random from the set of integers from 2 to 9 inclusive, what is the probability that 96a/b is the square of an integer?

Let k^2 = 96a/b = 2^5*3a/b

a/b = {2/3, 3/2, 4/6, 6/4, 6/9, 9/6, 3/8, 8/3} : 8 pairs

Total ways to select a & b = 8*8 = 64

Favorable ways to select a & b = 4

The probability that 96a/b is the square of an integer = 8/64 = 1/8

IMO B

Regor60 · Answer

The standard interpretation unless indicated otherwise when members are selected from a set is that they are selected without replacement, meaning that the same number cannot be selected twice.

kevincan · Answer

If two people each choose a prime number less than 20, is it possible that they choose the same number ?

If an and b are chosen at random from the set of integers from 2 to 9

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