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04 Mar 2026, 11:57
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C
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Question Stats:
56% (02:32) correct
44%
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based on 57
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If two distinct integers are chosen at random from the integers from 1 to 10 inclusive, what is the probability that the product of the two integers is a multiple of 8?
A. 1/9
B. 2/9
C. 4/15
D. 1/3
E. 2/5
A. 1/9
B. 2/9
C. 4/15
D. 1/3
E. 2/5
04 Mar 2026, 22:13
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kevincan
Distinct integers from numbers 1 to 10 (both inclusive). But are multiples of 8.
The possible numbers are (1,8);(2,8);(3,8);(4,8);(5,8);(6,8);(7,8);(9,8);(10,8). So totally 9 values.
Multiples of 8 , which are from 4 as one number.
(4 , * multiple of 2 ) = (4,2); (4,6);
So, we get 3 values.
Total values = 9+3 =12
Probability = 12/(10C2) = (4/15)
Option C
04 Mar 2026, 23:17
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Key concept being tested: Probability using combinations — specifically, systematic case enumeration for divisibility.
The most common mistake here is trying to use a formula instead of just listing the cases. There is no elegant shortcut for "multiples of 8" — you need to be methodical.
Step 1 — Total outcomes.
Choosing 2 distinct integers from 1 to 10: C(10, 2) = 45 pairs. This is your denominator.
Step 2 — Identify what "multiple of 8" requires.
8 = 2^3, so the product needs at least three factors of 2. That means you need pairs where the combined prime factorization includes 2^3.
Even numbers in 1-10 and their powers of 2:
- 2 = 2^1
- 4 = 2^2
- 6 = 2^1 x 3
- 8 = 2^3
- 10 = 2^1 x 5
Step 3 — Systematically find all valid pairs.
Pairs containing 8 (8 = 2^3 already, so any partner works):
(8,1), (8,2), (8,3), (8,4), (8,5), (8,6), (8,7), (8,9), (8,10) → 9 pairs
Pairs NOT containing 8 (need combined powers of 2 ≥ 3):
- (4, 2): 2^2 x 2^1 = 2^3 ✓
- (4, 6): 2^2 x 2^1 = 2^3 ✓
- (4, 10): 2^2 x 2^1 = 2^3 ✓
- (2, 6): 2^1 x 2^1 = 2^2 ✗
- (2, 10): 2^1 x 2^1 = 2^2 ✗
- (6, 10): 2^1 x 2^1 = 2^2 ✗
→ 3 additional pairs
Step 4 — Calculate probability.
Favorable pairs = 9 + 3 = 12
P = 12/45 = 4/15 → Answer: C
Common trap: Students forget the (4,6) and (4,10) pairs — they fixate on 8 as the only source of the "x8" condition and miss that 4 x any multiple of 2 also clears the threshold. Always map out the even numbers' prime factorizations before counting.
Takeaway: For "divisible by n" probability questions, build a systematic list by fixing the element with the highest power of the prime factor first, then check remaining cases — don't try to count by formula alone.
— Kavya | 725 (99th percentile), GMAT Focus
The most common mistake here is trying to use a formula instead of just listing the cases. There is no elegant shortcut for "multiples of 8" — you need to be methodical.
Step 1 — Total outcomes.
Choosing 2 distinct integers from 1 to 10: C(10, 2) = 45 pairs. This is your denominator.
Step 2 — Identify what "multiple of 8" requires.
8 = 2^3, so the product needs at least three factors of 2. That means you need pairs where the combined prime factorization includes 2^3.
Even numbers in 1-10 and their powers of 2:
- 2 = 2^1
- 4 = 2^2
- 6 = 2^1 x 3
- 8 = 2^3
- 10 = 2^1 x 5
Step 3 — Systematically find all valid pairs.
Pairs containing 8 (8 = 2^3 already, so any partner works):
(8,1), (8,2), (8,3), (8,4), (8,5), (8,6), (8,7), (8,9), (8,10) → 9 pairs
Pairs NOT containing 8 (need combined powers of 2 ≥ 3):
- (4, 2): 2^2 x 2^1 = 2^3 ✓
- (4, 6): 2^2 x 2^1 = 2^3 ✓
- (4, 10): 2^2 x 2^1 = 2^3 ✓
- (2, 6): 2^1 x 2^1 = 2^2 ✗
- (2, 10): 2^1 x 2^1 = 2^2 ✗
- (6, 10): 2^1 x 2^1 = 2^2 ✗
→ 3 additional pairs
Step 4 — Calculate probability.
Favorable pairs = 9 + 3 = 12
P = 12/45 = 4/15 → Answer: C
Common trap: Students forget the (4,6) and (4,10) pairs — they fixate on 8 as the only source of the "x8" condition and miss that 4 x any multiple of 2 also clears the threshold. Always map out the even numbers' prime factorizations before counting.
Takeaway: For "divisible by n" probability questions, build a systematic list by fixing the element with the highest power of the prime factor first, then check remaining cases — don't try to count by formula alone.
— Kavya | 725 (99th percentile), GMAT Focus
