Bob likes 2 digit numbers n>10 such that both digits are squares. For example, 10 and 41 are two such numbers. How many of these numbers are there?

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Answer 1 · 2016-10-30T23:47:37

There are $8$ $8$ numbers that fit this criteria.

Let's list the single digit numbers that are perfect squares.

There is $0, 1, 4 and 9$ $0, 1, 4 and 9$ .

We need to pick $2$ $2$ numbers out of these four numbers. Order does matter, so we use the permutation formula.

$number of combinations = \frac{n!}{(n - r)!}$ $"number of combinations" = (n!)/((n - r)!)$

$number of combinations = \frac{4!}{(4 - 2)!}$ $"number of combinations" = (4!)/((4 - 2)!)$

$number of combinations = \frac{24}{2}$ $"number of combinations" = 24/2$

$number of combinations = 12$ $"number of combinations" = 12$

However, we cannot have $n < 10$ $n<10$ , so we have to remove $09, 01, 04, 00$ $09, 01, 04, 00$ , which leaves us with 8 possible numbers.