Question

If the numbers between 1 to 65 which will be divisible by 4 are taken and then if the number present in the units place and tens place is swapped, post which they are written in ascending order, then which of the following number will be at 10th place from the last?

• A. 40
• B. 24
• C. 44
• D. 25

Hint:

Swap the units and tens digit of each multiple of 4, then sort ascending.

Answer 1

The Correct Option is A

Step 1: Notice the units digits of multiples of 4 repeat in a cycle.
For 4, 8, 12, 16, 20, 24, and so on, the units digit cycles as $4, 8, 2, 6, 0$ and then repeats. Since swapping units and tens turns this units digit into the new tens digit, the swapped numbers fall into 5 tens-digit brackets: 0, 2, 4, 6 and 8.

Step 2: Count how many of the 16 numbers fall in each bracket.
Multiples with units digit 0: 20, 40, 60 (3 numbers). Units digit 2: 12, 32, 52 (3 numbers). Units digit 4: 4, 24, 44, 64 (4 numbers). Units digit 6: 16, 36, 56 (3 numbers). Units digit 8: 8, 28, 48 (3 numbers). Total: $3+3+4+3+3=16$, which checks out.

Step 3: Work backward from the largest bracket to count off 10 places.
In ascending order, the brackets appear as 0, 2, 4, 6, 8 (smallest new tens digit first). Counting from the last number backward: the 8-bracket covers the 1st to 3rd from the last, the 6-bracket covers the 4th to 6th from the last, and the 4-bracket covers the 7th to 10th from the last.

Step 4: Read off the number that lands on the 10th spot.
The 4-bracket, sorted ascending, is 40, 42, 44, 46 (from swapping 4, 24, 44, 64). The 10th from the last is the smallest of these four, which is 40.

Final Answer:
Counting backward in brackets instead of listing every swapped number one by one, we again land on 40 as the 10th number from the last.
\[ \boxed{40} \]