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There are 17 such numbers.

Q: How many of the numbers from 10 through 98 have the sum of their digits equal to a perfect and 8203 square?

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14

102 = 100 which is the first possible three digit number that is a perfect square. 312 = 961 which is the last possible three digit number that is a perfect square. So there are 22 three digit positive numbers that are perfect squares.

There are 11 numbers from 10 to 20 if we include 10 and 20. Of those, only 16 is a perfect square. So the probability if 1/11 or .09 repeating.

121 because 2x1x1=2 and 11x11=121

To any set that contains it! It belongs to {-3.5}, or {-3.5, sqrt(2), pi, -3/7}, or all numbers between -43 and 53, or multiples of 0.5, or rational numbers, or real numbers, or square roots of 12.25, or complex numbers, etc.

Related questions

16 of them.

18

14

17 of them. 17 of them. 17 of them. 17 of them.

69

The sum of the digits in '89' is 17, so I don't need toworry about any perfect square bigger than 16.1: 104: 13, 22, 31, 409: 18, 27, 36, 45, 54, 63, 72, 8116: 79That's 14 of them .

If the number with the digits reversed can have a leading 0 so that it is a 1-digit number, then 16. Otherwise 13.

121.

the highest sum of the numbers is 17 and the lowest is 1. The only perfect squares in that range are 1,4,9, and 16. That means the following numbers will work: 10,13.18,22,27,31,36,40,45,54,63,72,79,81,88, and 90; that is 16 numbers

There are 14 such numbers: 10, 13, 18, 22, 27, 31, 36, 40, 45, 54, 63, 72, 79, 81

8362 = 698896

No. Square numbers (or perfect square numbers) are squares of integers. The perfect square numbers are 1, 4, 9, 16, 25, 36 ... 20 is not in the series.