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

Jul 19, 2016

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 $11 + 39 + 61 + 89 = 200$

Jul 19, 2016

$200$

#### 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 $100 {a}^{2} + 20 a + 1$ 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 $100 {b}^{2} - 20 b + 1$ 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

$11 + 39 + 61 + 89 = 200$