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


Answer:

55

Step by Step Explanation:
  1. Two-digit squares can only start with 1,2,3,4,6 or 8.1,2,3,4,6 or 8.
  2. The number starting with 11 goes 16491649
    Since the only two-digit square number starting with 11 is 16,16, the 22 -digit square number starting with 66 is 64,64, the square number starting with 44 is 4949. Now, it can't be continued further as no two-digit square number starts with 9.9.
    Therefore, the required integer starting with 11 is 1649.1649.
    The number starting with 22 goes 2525 and then can't be continued as no two-digit square number starts with 5.5.
    Therefore, the required integer starting with 22 is 25.25.
  3. Similarly, the required integer starting 33 is 3649.3649.
    The required integer starting 44 is 49.49.
    The required integer starting 66 is 649.649.
    The required integer starting 88 is 81649.81649.
    Observe that 8164981649 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 5.5.

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