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Q: How many three digit numbers exist in which the last digit is the same as the last digit of that number squared?

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Such a number does not exist.

Ordinarily, the square root of -70 does not exist since a positive number squared or a negative number squared results in a positive number. If complex numbers are permitted, then the square root of -70 is i√70 ~= 8.367i where i is the imaginary square root of -1 such that i2 = -1.

no, every number is a real number --- There are numbers that are not real numbers. They are called imaginary numbers, and have the property that when they are squared, the result is negative. The square root of -1 is called i, and the square root of any other negative number is i times the square root of the absolute value of the number. So the square root of -4 is 2i.

Such a number does not exist.

It is best to do this on a calculator. If you want to do it without one, you can try different numbers, to see which number, when squared, is closest to 10. For example, 3 squared is 9, and 4 squared is 16, so the square root of 10 is obviously between those two. Faster methods exist, too.

There are an infinite number of prime numbers.

There is no way to determine the amount of numbers that exist. Unlike physical objects, numbers are not finite. There is literally no end to the amount of numbers that can exist.

Yes. A 17-digit number would be less than a quintillion.

agazillion gazillion do not exist, they aren't numbers.

I am sure that there are 25 prime numbers exist in mathematics

y is greater than 0 x exist in a set of real numbers

That number does not exist since numbers go on infinitely.

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