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

Step 1: Note the starting position and what needs to change.
The sets begin as Set A = 7, Set B = 28, Set C = 196, Set D = 34, Set E = 5. Set C already works with Set A and Set B, since \(196 = 7 \times 28\). It fails with Set D and Set E, since \(196 / 34\) and \(196 / 5\) are not whole numbers. So the swaps must fix Set D and Set E without breaking the good relationship with Set A and Set B.

Step 2: Make the first swap.
Swap the 7 in Set A with the 2 in Set B. Set A becomes 2 and Set B becomes 78. Set C is still 196, and \(196 / 2 = 98\), so Set A still divides Set C exactly with a new quotient. This move alone is not the full answer, but it puts a 2 in Set A that pairs usefully with the number formed later in Set C.

Step 3: Make the second swap.
Swap the 4 in Set D with the 5 in Set E. Set D becomes 35 and Set E becomes 4. Set C is still 196.

Step 4: Make the third swap and check every set.
Swap the 9 in Set C with the 5 that now sits in Set D. Set C becomes 156 and Set D becomes 39. Check all four sets against Set C = 156: \(156 / 2 = 78\) (Set A), \(156 / 78 = 2\) (Set B), \(156 / 39 = 4\) (Set D), \(156 / 4 = 39\) (Set E). Every one of them divides 156 exactly, so the condition is finally met.

Step 5: Confirm this is the minimum.
Two of the four sets, A and B, were already correct at the very start, but Set D and Set E both still needed a digit to move in and out of them, and a workable Set C could only be built once digits from Set C, Set D and Set E had each changed places once. There is no way to fix both Set D and Set E, and still keep Set C consistent with all four sets, using only 1 or 2 swaps, since at least three digits across Sets B, C, D and E needed to relocate before every division came out exact. Three swaps is the smallest number that gets there.

Final Answer:
The minimum number of rearrangements needed is 3. \[ \boxed{3} \]