How about 7*7 = 49 because 4+9 = 13
No. Think 13, 22, 31 for a start!
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 .
13
I did it by trial and error. 7^2 is 49. 4 + 9 = 13
Start with the smallest multiple of 13 and continue with the next smallest until finding one that fits the specifications. 13, sum of digits is 1 + 3 = 4 which is not prime 26, sum of digits is 2 + 6 = 8 which is not prime 39, sum of digits is 3 + 9 = 12 which is not prime 52, sum of digits is 5 + 2 = 7 which is prime So, 52 is the smallest positive multiple of 13 for which the sum of its digits is prime.
13
If the number with the digits reversed can have a leading 0 so that it is a 1-digit number, then 16. Otherwise 13.
65
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.
65 has 13 as a factor because 13 divides into 65 evenly with no remainder. The sum of the digits of 65 is 11 because 6 + 5 = 11
65
1 and 12