Question:medium

There are five sets of digits, Set A, Set B, Set C, Set D and Set E, arranged in a row as shown below. Set A holds one digit, Set B holds two digits, Set C holds three digits, Set D holds two digits and Set E holds one digit.

Set A: 7, Set B: 28, Set C: 196, Set D: 34, Set E: 5.

A rearrangement means picking one digit out of one set and swapping it with one digit from a different set. The goal is to keep making such swaps, one at a time, until the three-digit number in Set C becomes an exact multiple of the numbers formed by every other set, that is, of Set A, Set B, Set D and Set E, all at once. In the starting arrangement above, Set C (196) is already a multiple of Set A (77) and of Set B (28), since \(196 = 7 \times 28\), but it is not a multiple of Set D (34) or of Set E (55).
After the digits are rearranged so that Set C becomes a multiple of Set A, Set B, Set D and Set E all together, which pair of digits ends up in Set A and Set E?

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Use the same rearranged sets you found while working out the minimum-swaps question, and read off the single digit sitting in Set A and the single digit sitting in Set E.
Updated On: Jul 10, 2026
  • 2 and 4
  • 2 and 6
  • 3 and 6
  • 3 and 9
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The Correct Option is A

Solution and Explanation

Rather than re-deriving the whole rearrangement from scratch, check each candidate pair against the requirement that whatever single digits sit in Set A and Set E must both divide the final 3-digit number in Set C exactly, since Set C has to be a multiple of every other set, including A and E on their own.

  1. 2 and 4: a three-digit multiple of both 2 and 4, such as 156, is easy to build from the nine digits available, and this pair matches a working final layout, Set A = 2, Set B = 78, Set C = 156, Set D = 39, Set E = 4, where every set divides 156 exactly.
  2. 2 and 6: this would need Set C to be divisible by both 2 and 6 at once, but forcing a 6 into Set E leaves no way to arrange the rest of the digits so that Set B and Set D also stay valid two-digit divisors of the same Set C, so this pair cannot be completed into a full working solution.
  3. 3 and 6: the same problem shows up here too: forcing a 6 into Set E blocks Set B and Set D from remaining exact two-digit divisors of a three-digit Set C, so this pair also fails.
  4. 3 and 9: putting 9 in Set E means it can no longer sit inside Set C, and without it Set C cannot be rebuilt as a three-digit multiple of every remaining set from the digits left over, so this pair fails too.

Only the pair 2 and 4 can actually be extended into a complete, valid rearrangement of all nine digits where Set C is a multiple of Set A, Set B, Set D and Set E at the same time.

Let's summarize:

  • Set A and Set E must each hold a single digit that exactly divides the final Set C.
  • Only the pair 2 and 4 can be completed into a full nine-digit rearrangement that satisfies every set at once.
  • The working layout is Set A = 2, Set B = 78, Set C = 156, Set D = 39, Set E = 4.

So Set A and Set E hold the digits 2 and 4.

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