Question

What is the sum of all two-digit whole numbers whose squares end with the digits 21?

Answer 1

200

Explanation:

A square number ending in a '1' can only be produced by squaring a number ending in a '1' or a '9'. Source. This helps a lot in the search. Quick bit of number crunching gives:

from our table we can see that

`11^2 = 121`

`39^2 = 1521`

`61^2 = 3721`

`89^2 = 7921`

So `11+39+61+89 = 200`

Answer 2

`200`

Explanation:

If the last digits of a square of a two digit number are `21`, unit's digit is either `1` or `9`.

Now, if tens digit is `a` and units digit is `1`, it is of type `100a^2+20a+1` and we can have last two digits as `21` if `a` is `1` or `6` i.e. numbers are `10+1=11` and `60+1=61`.

If ten's digit is `b` and unit digit is `9`, it is of type `100b^2-20b+1` and we can have last two digits as `21` if `b` is `4` or `9` i.e. numbers are `40-1=39` and `90-1=89`.

Hence, sum of all such two digit numbers is

`11+39+61+89=200`