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15 Mar 2026, 10:16
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C
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Difficulty:
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Question Stats:
6% (03:21) correct
94%
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wrong
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Let S be the set of six-digit integers formed from the digits 5, 6, 7, 8, and 9, each appearing at least once. What is the approximate sum of all numbers in S?
A. 300,000,000
B. 900,000,000
C. 1,400,000,000
D. 1,800,000,000
E. 2,800,000,000
A. 300,000,000
B. 900,000,000
C. 1,400,000,000
D. 1,800,000,000
E. 2,800,000,000
15 Mar 2026, 11:03
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So this is a classic "sum of all arrangements" problem, and the trick is that you don't actually need to list out every single number. Everyone's instinct is to try that and get buried in calculation.
Here's the move: we're making six-digit numbers using 5, 6, 7, 8, 9 where each digit appears at least once. Since we have 5 different digits and need 6 positions, exactly one digit appears twice.
How many numbers in S?
- Pick which digit appears twice: 5 ways
- Arrange those 6 digits: 6!/2! = 360 arrangements per digit
- Total: 5 × 360 = 1,800 numbers
What's the sum?
Here's the elegant part. Think about symmetry. In all 1,800 numbers, how many times does each digit appear in each position?
Total digit placements = 1,800 numbers × 6 positions = 10,800
Each of the 5 digits appears 10,800 / 5 = 2,160 times across all positions.
By symmetry, each digit appears equally in each place value:
2,160 / 6 positions = 360 times per position per digit
Sum in each position = 360 × (5 + 6 + 7 + 8 + 9) = 360 × 35
Contribution to total:
360 × 35 × (100,000 + 10,000 + 1,000 + 100 + 10 + 1)
= 360 × 35 × 111,111
= 12,600 × 111,111
≈ 1,400,000,000
Answer: C
The symmetry insight is what saves you here—you avoid computing 1,800 different sums.
Here's the move: we're making six-digit numbers using 5, 6, 7, 8, 9 where each digit appears at least once. Since we have 5 different digits and need 6 positions, exactly one digit appears twice.
How many numbers in S?
- Pick which digit appears twice: 5 ways
- Arrange those 6 digits: 6!/2! = 360 arrangements per digit
- Total: 5 × 360 = 1,800 numbers
What's the sum?
Here's the elegant part. Think about symmetry. In all 1,800 numbers, how many times does each digit appear in each position?
Total digit placements = 1,800 numbers × 6 positions = 10,800
Each of the 5 digits appears 10,800 / 5 = 2,160 times across all positions.
By symmetry, each digit appears equally in each place value:
2,160 / 6 positions = 360 times per position per digit
Sum in each position = 360 × (5 + 6 + 7 + 8 + 9) = 360 × 35
Contribution to total:
360 × 35 × (100,000 + 10,000 + 1,000 + 100 + 10 + 1)
= 360 × 35 × 111,111
= 12,600 × 111,111
≈ 1,400,000,000
Answer: C
The symmetry insight is what saves you here—you avoid computing 1,800 different sums.
16 Mar 2026, 01:46
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This is a clever counting + symmetry problem. Let me walk through it step by step.
Step 1: Count how many numbers are in set S.
We have 6 positions and 5 digits (5, 6, 7, 8, 9), each appearing at least once. So exactly 1 digit gets repeated.
- Choose which digit repeats: 5 ways
- Arrange the 6 digits (with one pair of identical digits): 6!/2! = 360 ways
- Total numbers in S: 5 × 360 = 1,800
Step 2: Figure out how often each digit appears in each position.
Key Insight — by symmetry, no digit is "favored" in any position.
- Total digit slots across all numbers: 1,800 × 6 = 10,800
- Each of the 5 digits appears: 10,800 / 5 = 2,160 times total
- Spread equally across 6 positions: 2,160 / 6 = 360 times per position
So in the hundreds-thousands place, the digit 5 appears 360 times, the digit 6 appears 360 times, and so on for every position.
Step 3: Calculate the total sum.
For any single position, the sum of all digits placed there is:
360 × (5 + 6 + 7 + 8 + 9) = 360 × 35 = 12,600
Now multiply by the place values. The ones place contributes 12,600 × 1, tens place contributes 12,600 × 10, and so on up to 12,600 × 100,000.
Total = 12,600 × (1 + 10 + 100 + 1,000 + 10,000 + 100,000)
Total = 12,600 × 111,111 ≈ 1,400,000,000
Answer: C
The big takeaway: when a problem asks for the sum of ALL numbers in a symmetric set, use the fact that each digit appears equally often in each position. Then just compute (appearances per position) × (sum of digits) × (sum of place values).
Step 1: Count how many numbers are in set S.
We have 6 positions and 5 digits (5, 6, 7, 8, 9), each appearing at least once. So exactly 1 digit gets repeated.
- Choose which digit repeats: 5 ways
- Arrange the 6 digits (with one pair of identical digits): 6!/2! = 360 ways
- Total numbers in S: 5 × 360 = 1,800
Step 2: Figure out how often each digit appears in each position.
Key Insight — by symmetry, no digit is "favored" in any position.
- Total digit slots across all numbers: 1,800 × 6 = 10,800
- Each of the 5 digits appears: 10,800 / 5 = 2,160 times total
- Spread equally across 6 positions: 2,160 / 6 = 360 times per position
So in the hundreds-thousands place, the digit 5 appears 360 times, the digit 6 appears 360 times, and so on for every position.
Step 3: Calculate the total sum.
For any single position, the sum of all digits placed there is:
360 × (5 + 6 + 7 + 8 + 9) = 360 × 35 = 12,600
Now multiply by the place values. The ones place contributes 12,600 × 1, tens place contributes 12,600 × 10, and so on up to 12,600 × 100,000.
Total = 12,600 × (1 + 10 + 100 + 1,000 + 10,000 + 100,000)
Total = 12,600 × 111,111 ≈ 1,400,000,000
Answer: C
The big takeaway: when a problem asks for the sum of ALL numbers in a symmetric set, use the fact that each digit appears equally often in each position. Then just compute (appearances per position) × (sum of digits) × (sum of place values).
