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Probability is one of the most important GMAT Quant topics because it often combines logic, counting, set theory, and permutations & combinations. Many students try to solve probability questions by listing every possible case, but GMAT probability...- Jun 10
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27 May 2026, 21:20
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
22% (02:36) correct
78%
(02:48)
wrong
based on 18
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If x and y are integers chosen independently and at random from the first ten positive integers, what’s the probability that \(x^3y - xy^3\) is a multiple of 5?
A. \(\frac{8}{25}\)
B. \(\frac{2}{5}\)
C. \(\frac{12}{25}\)
D. \(\frac{3}{5}\)
E. \(\frac{17}{25}\)
A. \(\frac{8}{25}\)
B. \(\frac{2}{5}\)
C. \(\frac{12}{25}\)
D. \(\frac{3}{5}\)
E. \(\frac{17}{25}\)
29 May 2026, 22:58
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The factoring step is right but the solution above misses something important. The expression is xy(x-y)(x+y), so the product is divisible by 5 if ANY of the four factors — x, y, (x-y), or (x+y) — is divisible by 5. The reply above only checks whether x or y is a multiple of 5, forgetting about the (x-y) and (x+y) terms.
Concept: Probability with modular arithmetic (Multiples and Factors).
1. Factor: x^3*y - x*y^3 = xy(x^2 - y^2) = xy(x-y)(x+y). Good start.
2. Work mod 5. The integers 1-10 have residues: {0,0,1,1,2,2,3,3,4,4} corresponding to {5,10,1,6,2,7,3,8,4,9}. Each residue 0-4 appears exactly twice.
3. The product fails to be divisible by 5 only when ALL of these hold simultaneously: x not divisible by 5, y not divisible by 5, (x-y) not divisible by 5, (x+y) not divisible by 5.
4. Let a = x mod 5 and b = y mod 5. We need: a ≠ 0, b ≠ 0, a ≠ b, and a + b ≠ 0 (mod 5).
5. For each value of a in {1,2,3,4}, count valid b values in {1,2,3,4} where b ≠ a and b ≠ 5-a:
- a=1: excluded b = 1 and 4. Valid b: {2,3}. Two options.
- a=2: excluded b = 2 and 3. Valid b: {1,4}. Two options.
- a=3: excluded b = 3 and 2. Valid b: {1,4}. Two options.
- a=4: excluded b = 4 and 1. Valid b: {2,3}. Two options.
6. Each residue class has 2 integers out of 10. So each (a,b) residue pair has probability (2/10)(2/10) = 4/100. There are 4 * 2 = 8 valid "failure" pairs, so P(NOT divisible by 5) = 8 * (4/100) = 32/100 = 8/25.
7. P(divisible by 5) = 1 - 8/25 = 17/25.
Answer: E
The 9/25 answer only accounts for x or y being a multiple of 5. It ignores pairs like (1,4), where x+y = 5. I got burned by a similar trap the first time I saw this type — factoring is never the last step, you have to think about all the factors that could hit zero mod 5.
Concept: Probability with modular arithmetic (Multiples and Factors).
1. Factor: x^3*y - x*y^3 = xy(x^2 - y^2) = xy(x-y)(x+y). Good start.
2. Work mod 5. The integers 1-10 have residues: {0,0,1,1,2,2,3,3,4,4} corresponding to {5,10,1,6,2,7,3,8,4,9}. Each residue 0-4 appears exactly twice.
3. The product fails to be divisible by 5 only when ALL of these hold simultaneously: x not divisible by 5, y not divisible by 5, (x-y) not divisible by 5, (x+y) not divisible by 5.
4. Let a = x mod 5 and b = y mod 5. We need: a ≠ 0, b ≠ 0, a ≠ b, and a + b ≠ 0 (mod 5).
5. For each value of a in {1,2,3,4}, count valid b values in {1,2,3,4} where b ≠ a and b ≠ 5-a:
- a=1: excluded b = 1 and 4. Valid b: {2,3}. Two options.
- a=2: excluded b = 2 and 3. Valid b: {1,4}. Two options.
- a=3: excluded b = 3 and 2. Valid b: {1,4}. Two options.
- a=4: excluded b = 4 and 1. Valid b: {2,3}. Two options.
6. Each residue class has 2 integers out of 10. So each (a,b) residue pair has probability (2/10)(2/10) = 4/100. There are 4 * 2 = 8 valid "failure" pairs, so P(NOT divisible by 5) = 8 * (4/100) = 32/100 = 8/25.
7. P(divisible by 5) = 1 - 8/25 = 17/25.
Answer: E
The 9/25 answer only accounts for x or y being a multiple of 5. It ignores pairs like (1,4), where x+y = 5. I got burned by a similar trap the first time I saw this type — factoring is never the last step, you have to think about all the factors that could hit zero mod 5.
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