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25 Jun 2025, 23:55
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sahitiyalavarthi
“Distinct non-zero digits” means the digits must all be different from each other and none of them can be 0. So, for example, 123 is fine, but 121 is not (repeated digit), and 120 is not (includes 0).
13 Oct 2025, 09:45
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This is a classic counting problem where you need to juggle multiple constraints simultaneously. Let me walk you through how to approach this systematically.
Understanding What We Need:
We're looking for 3-digit numbers (let's call them ABC where A is hundreds, B is tens, C is units) that satisfy ALL of these:
The Key Insight:
Notice how A only has 3 possible values? That's your most restrictive constraint, so let's organize our counting around it. Here's what you need to see: when A is even (specifically 8), it affects how many choices we have for C!
Let's Count Case by Case:
Case 1: When A = 7
Case 2: When A = 8
Case 3: When A = 9
Final Count: \(28 + 21 + 28 = 77\)
Answer: E
The complete solution on Neuron reveals the systematic framework for identifying constraint hierarchies in any counting problem, plus time-saving techniques for recognizing similar patterns across different question types. You can check out the step-by-step solution on Neuron by e-GMAT to master the constraint-based counting approach systematically. You can also explore other GMAT official questions with detailed solutions on Neuron for structured practice here.
Hope this helps!
Understanding What We Need:
We're looking for 3-digit numbers (let's call them ABC where A is hundreds, B is tens, C is units) that satisfy ALL of these:
- Greater than 700, so A must be 7, 8, or 9
- Even numbers, so C must be 2, 4, 6, or 8
- All digits distinct (no repeats)
- All digits non-zero (only use 1-9)
The Key Insight:
Notice how A only has 3 possible values? That's your most restrictive constraint, so let's organize our counting around it. Here's what you need to see: when A is even (specifically 8), it affects how many choices we have for C!
Let's Count Case by Case:
Case 1: When A = 7
- C must be even and different from 7, so C can be: 2, 4, 6, or 8 → that's 4 choices
- B must be different from both 7 and whatever C is → from 9 digits, exclude 2 (A and C) → 7 choices
- Total: \(1 \times 7 \times 4 = 28\)
Case 2: When A = 8
- Here's the critical part: C must be even AND different from 8
- So C can only be: 2, 4, or 6 → that's just 3 choices (not 4!)
- B must be different from 8 and C → still 7 choices
- Total: \(1 \times 7 \times 3 = 21\)
Case 3: When A = 9
- C must be even and different from 9 → C can be: 2, 4, 6, or 8 → 4 choices
- B different from 9 and C → 7 choices
- Total: \(1 \times 7 \times 4 = 28\)
Final Count: \(28 + 21 + 28 = 77\)
Answer: E
The complete solution on Neuron reveals the systematic framework for identifying constraint hierarchies in any counting problem, plus time-saving techniques for recognizing similar patterns across different question types. You can check out the step-by-step solution on Neuron by e-GMAT to master the constraint-based counting approach systematically. You can also explore other GMAT official questions with detailed solutions on Neuron for structured practice here.
Hope this helps!
23 Feb 2026, 08:18
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To find: The number of even positive three-digit integers greater than 700 with distinct nonzero digits.
Solution:
Let's first understand the requirements - the constraints on the digits based on the kind of numbers we want to create:
Since the number of choices for the last digit depend on the first digit being even/odd, we will solve this question in two cases:
Case 1: Numbers with hundreds’ digit = 7 or 9
Case 2: Numbers with hundreds’ digit = 8
Considering both cases, total number of 3-digit even numbers greater than 700 with distinct nonzero digits = 28 + 21 + 28 = 77
Correct Answer: E
Shweta Koshija
GMAT, GRE, SAT Coach for 10+ years
Solution:
Let's first understand the requirements - the constraints on the digits based on the kind of numbers we want to create:
- Even number: This means we can only have 0, 2, 4, 6, or 8 in the units' place.
- Non-zero digits: We only have 9 digits overall, excluding 0. Specifically, the units' digit cannot be 0 and has only 4 choices (2, 4, 6, or 8)
- Greater than 700: The hundreds’ digit can only be 7 or 8 or 9.
- Distinct digits: We need 3 different digits to form our 3-digit numbers.
Since the number of choices for the last digit depend on the first digit being even/odd, we will solve this question in two cases:
Case 1: Numbers with hundreds’ digit = 7 or 9
- Number of choices for the hundreds digit = 2 (7 or 9)
- Number of choices for the units’ digit = 4 (2, 4, 6, or 8)
- Finally, number of choices for the tens’ digit = 10 – 1 – 1 – 1 = 7 (excluding 0, the units’ digit, the hundreds’ digit)
- So, total number of 3-digit numbers formed in this case = 2 * 4 * 7 = 56.
Case 2: Numbers with hundreds’ digit = 8
- Number of ways to fill the hundreds digit = 1 (only 8)
- Number of choices for the units’ digit = 3 (2, 4, or 6)
- Finally, number of choices for the tens’ digit = 10 – 1 – 1 – 1 = 7 (excluding 0, the units’ digit, the hundreds’ digit)
- So, total number of 3-digit numbers formed in this case = 1 * 3 * 7 = 21.
Considering both cases, total number of 3-digit even numbers greater than 700 with distinct nonzero digits = 28 + 21 + 28 = 77
Correct Answer: E
Shweta Koshija
GMAT, GRE, SAT Coach for 10+ years
