4, 400, 40000 etc
121, 12100, 1210000 etc
121.
11 squared is 121
No. Think 13, 22, 31 for a start!
How about 7*7 = 49 because 4+9 = 13
4
It is a factor of 32, which means it is one of 1, 2, 4, 8, 16, and 32. The perfect squares among the factors of 32 are 4 and 16. Of those two, the only one with a sum of digits that is odd is 16.
To find the numbers between 10 and 87 with the sum of their digits equal to a perfect square, we need to consider each number individually. The perfect squares between 1 and 9 are 1, 4, and 9. For each perfect square, we can list the numbers that meet the criteria: For the perfect square 1, the numbers are 10, 19, 28, 37, 46, 55, 64, 73, 82. For the perfect square 4, the numbers are 40, 49, 58, 67, 76, 85. For the perfect square 9, the numbers are 90. Therefore, there are 14 numbers between 10 and 87 that have the sum of their digits equal to a perfect square.
4*4=16 sum of 1+6 is 7, difference is 5 so 16 is answer
To find the two-digit numbers whose digits sum to a perfect square, we first identify the possible sums of the digits, which range from 1 (1+0) to 18 (9+9). The perfect squares in this range are 1, 4, 9, and 16. The valid two-digit combinations for these sums are: for 4 (14, 23, 32, 41, 50), for 9 (18, 27, 36, 45, 54, 63, 72, 81, 90), and for 16 (79). Counting all valid combinations gives us a total of 20 two-digit numbers.
16
4
Well, let's see. Perfect cubes that are two digits: 27 64 Could it be 27? Well, 2+7 is 9, and that's a perfect square with a square root of 3, and the cube root of 27 is three. Looks like we've found our answer, especially since 6+4 = 10, which is NOT a perfect square.