What is the sum of all 4-digit numbers less than 2000 formed by the digits 1, 2, 3 and 4, where none of the digits is repeated?
- A7998Correct
- B8028
- C8878
- D9238
Explanation
Here's a brief explanation:
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Identify the numbers: The numbers must be 4-digit, less than 2000, and formed by digits 1, 2, 3, 4 without repetition. This means the thousands digit must be 1. The remaining digits (2, 3, 4) can be arranged in the hundreds, tens, and units places. There are 3! = 6 such numbers.
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List the numbers: 1234, 1243, 1324, 1342, 1423, 1432.
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Calculate the sum by place value:
- Thousands place: All 6 numbers have 1 in the thousands place. Sum = 6 * 1000 = 6000.
- Hundreds place: The digits 2, 3, 4 each appear 2 times (3! / 3 = 2 times) in the hundreds place. Sum = (2+3+4) * 2 * 100 = 9 * 2 * 100 = 1800.
- Tens place: The digits 2, 3, 4 each appear 2 times in the tens place. Sum = (2+3+4) * 2 * 10 = 9 * 2 * 10 = 180.
- Units place: The digits 2, 3, 4 each appear 2 times in the units place. Sum = (2+3+4) * 2 * 1 = 9 * 2 * 1 = 18.
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Total Sum: Add the sums from each place value: 6000 + 1800 + 180 + 18 = 7998.
Therefore, option A is correct.
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