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).

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?

• A. 2 and 4
• B. 2 and 6
• C. 3 and 6
• D. 3 and 9

Hint:

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.

Answer 1

The Correct Option is A

Step 1: Recall the rearranged sets.
Solving the minimum-swaps question showed that swapping 7 (Set A) with 2 (Set B), then 4 (Set D) with 5 (Set E), then 9 (Set C) with 5 (Set D) gives the working arrangement Set A = 2, Set B = 78, Set C = 156, Set D = 39, Set E = 4.

Step 2: Confirm this arrangement is valid.
Check Set C = 156 against every other set: \(156/2=78\), \(156/78=2\), \(156/39=4\), \(156/4=39\). Each division comes out exact, so this is indeed a correct final arrangement, reached in the minimum 3 swaps found earlier.

Step 3: Read off Set A and Set E.
In this arrangement, Set A holds the single digit 2 and Set E holds the single digit 4.

Final Answer:
Set A and Set E hold the digits 2 and 4. \[ \boxed{2 \text{ and } 4} \]