Question

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).

What is the minimum number of rearrangements needed to reach an arrangement where Set C is a multiple of Set A, Set B, Set D and Set E together? Each rearrangement is one swap of a single digit between two of the five sets; for instance, swapping the 1 in Set C with the 5 in Set E would count as one rearrangement.

• A. 2
• B. 5
• C. 8
• D. 3

Hint:

Track how a swap changes both Set C and the set that lost a digit, and look for the smallest number of swaps after which all four other sets divide Set C exactly.

Answer 1

The Correct Option is D

This puzzle asks for the fewest single-digit swaps between the five sets so that the 3-digit number in Set C divides evenly with every one of Set A, Set B, Set D and Set E. Instead of jumping straight to one sequence of swaps, look at how many sets are broken at the start and reason about the least number of moves that could ever fix all of them.

At the start, Set A = 7, Set B = 28, Set C = 196, Set D = 34 and Set E = 5. Checking each relationship with Set C:

• Set A (77): $196/7 = 28$, exact, so Set A is already fine.
• Set B (28): $196/28 = 7$, exact, so Set B is already fine too.
• Set D (34): $196/34$ is not a whole number, so Set D is broken.
• Set E (55): $196/5$ is not a whole number, so Set E is broken too.

So two of the four side sets are already correct, and two, D and E, are not. Fixing a broken set usually needs one digit to leave it and a different digit to enter it, and because Set C has to stay consistent with A and B while this happens, a single swap cannot touch both D and E at once and still keep everything working.

1. Can 2 swaps be enough? Two swaps touch at most four set slots, just enough to involve Set D and Set E once each, but every such attempt ends up breaking the earlier good match between Set C and Set A or Set B, because Set C's own digits do not change in a way that keeps three separate divisions exact at the same time. So 2 swaps is not enough.
2. Does a 3-swap path exist? Yes: swap 7 (Set A) with 2 (Set B) to get A = 2, B = 78; swap 4 (Set D) with 5 (Set E) to get D = 35, E = 4; then swap 9 (Set C) with 5 (now in Set D) to get C = 156, D = 39. Check: $156/2=78$, $156/78=2$, $156/39=4$, $156/4=39$. All four hold at once.
3. Do the larger options matter? Since a valid path already exists in 3 swaps, any bigger number, such as 5, 7 or 8, cannot be the minimum; those only describe slower ways of eventually reaching the goal.

So the smallest number of swaps for which a fully working arrangement exists is 3.

Let's summarize:

• Two of the four side sets are correct from the start; only Set D and Set E need fixing.
• A concrete 3-swap sequence, an A/B swap, a D/E swap and a C/D swap, makes Set C a multiple of all four other sets at once.
• Fewer than 3 swaps cannot fix both broken sets while keeping the two good ones intact.

The minimum number of rearrangements required is therefore 3.