How many digits will be there in the largest integer for which e | Math Question and Answer

Answer:

$5$

Step by Step Explanation:

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 $5.$

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