How many digits will be there in the largest integer for which each pair of consecutive digits is a square^@?^@


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Answer:

^@ 5 ^@

Step by Step Explanation:
  1. Two-digit squares can only start with ^@ 1, 2, 3, 4, 6 \text{ or } 8. ^@
  2. 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. ^@
  3. 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.
  4. Hence, the number of digits in the largest integer for which each pair of consecutive digits is a square is ^@ 5. ^@

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