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

Answer 1

200

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#

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Answer 2

Eva Abramson

#200#

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#

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Answer 3

Eva Adamson

The sum of all two-digit whole numbers whose squares end with the digits 21 is 330.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.