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 \text{ or } 8. ^@
- The number starting with ^@ 1 ^@ goes ^@ 1 \rightarrow 6 \rightarrow 4 \rightarrow 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 \rightarrow 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. ^@
