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let nn be the largest positive integer with the following property: reading from left to right, each pair of consecutive digits of nn forms a perfect square. what are the leftmost three digits of nn?

Respuesta :

The leftmost three digits of nn would be 816.

Solution:

The two-digit perfect squares are $16, 25, 36, 49, 64, 81$. We try making a sequence starting with each one:

  • 16 - 64 - 49. This terminates since none of them end in a 9, giving us 1649.
  • 25
  • 36 - 64 - 49, 3649.
  • 49
  • 64 - 49, 649.
  • 81 - 16 - 64 - 49, 81649.

The largest is 81649, so our answer is [816].

The leftmost three digits of nn would be 816.

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