Discussion
Let S be the set of all 3-digit numbers containing the digits 3, 5 and 7 without repetition. The sum of all numbers in S is
Answer: A
The sum is 3330.
The digits 3, 5 and 7 without repetition make \(3! = 6\) numbers: 357, 375, 537, 573, 735 and 753. Each digit stands in each place (hundreds, tens, units) exactly twice.
\[ \begin{aligned} \text{sum} &= 2(3 + 5 + 7)(100 + 10 + 1) \\ &= 2 \times 15 \times 111 = 3330 \end{aligned} \]
Adding the six numbers directly gives the same total. Most options are multiples of 111, so the factor in front decides it: six numbers, with each digit appearing twice in each place, give \(2 \times 15 = 30\) lots of 111.
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