4
x64 is not a perfect square any number if it is a 3-digit perfect square and ending with 4 it can be 144,324 ,484 and 784 because the number is ending with 4 but the tens digit is not matching to any of the option so it is not a perfect square
I am pretty sure you can figure this out on your own. Raise different numbers to the square, until you get a 4-digit result. Similary, calculate the cube of different numbers, until you get a 4-digit number. If you want the SAME number to be both a perfect square and a perfect cube, then it must be a power of 6. In that case, just experiment raising different numbers to the sixth power, until you get a 4-digit number.
11 squared is 121
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.
10000 is the square of 100 so you're looking for the square of 99 which is 9801.
x64 is not a perfect square any number if it is a 3-digit perfect square and ending with 4 it can be 144,324 ,484 and 784 because the number is ending with 4 but the tens digit is not matching to any of the option so it is not a perfect square
9801
121.
992 = 9,801
It's 4. (31,622)2 = 999,950,884 .
I am pretty sure you can figure this out on your own. Raise different numbers to the square, until you get a 4-digit result. Similary, calculate the cube of different numbers, until you get a 4-digit number. If you want the SAME number to be both a perfect square and a perfect cube, then it must be a power of 6. In that case, just experiment raising different numbers to the sixth power, until you get a 4-digit number.
11 squared is 121
1024, square of 32.
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.
What is the sum of greatest 3-digit 4-digit 5digit
4,096 = (64)2 = (16)3
99*99=9801