what is the sum of all two-digit positive integers whose squares end with the digits 01?

Question

all about integers.....

Answer 1

Let the 2 digit number by xy = 10x + y

Square of this = 100x^2 + y^2 + 20xy

Now for the number to end in 01, y is either 1 or 9

also, 100x^2 will have 00 in the end (last 2 digits)

20xy will have a0 in the end a is an even integer because the multiplication is with 2

So, for y = 1,
you have

20xy must end in 00

so, 20x1 = 100 or 200 or 300....

20x = 100 ---- x = 5
or 20x = 200...x = 10 but we are looking for only 2 digit numbers so, the only possible number is 51

Now for y = 9, y^2 ends in 81

so, 20xy must end in 20

since this number has to be a multiple of 9 (because y = 9)
we have 180x must end in 20

since, 18 x 4 = 72
18 x 9 = 172
so possible values of x are 4 and 9

so numbers are 49 and 99

so sum of all numbers is

49 + 99 + 51 = 199

Answer 2

49 + 51 + 99
= 199

Answer 3

zero - none of them end with "01".