How many digits will be there in the largest integer for which each pair of consecutive digits is a square?
Answer:
5
- Two-digit squares can only start with 1,2,3,4,6 or 8.
- The number starting with 1 goes 1→6→4→9
Since the only two-digit square number starting with 1 is 16, the 2 -digit square number starting with 6 is 64, the square number starting with 4 is 49. Now, it can't be continued further as no two-digit square number starts with 9.
Therefore, the required integer starting with 1 is 1649.
The number starting with 2 goes 2→5 and then can't be continued as no two-digit square number starts with 5.
Therefore, the required integer starting with 2 is 25. - Similarly, the required integer starting 3 is 3649.
The required integer starting 4 is 49.
The required integer starting 6 is 649.
The required integer starting 8 is 81649.
Observe that 81649 is the largest integer for which each pair of censecutive digits is a square. - Hence, the number of digits in the largest integer for which each pair of consecutive digits is a square is 5.