The primary equation is N = 100a + 10b + c for integers a, b and c. Also, 100a + 10b + c = 11m where m = a^2 + b^2 + c^2. Substitution and simplification yield 100a + 10b + c = 11a^2 + 11b^2 + 11c^2. There are two cases, where b = a + c, or b = a + c - 11. Solving these two cases results in the answer of N = 550, 803.
It has to be something like 11*50=550 (5^2)+(5^2)=50 Therefore, 550 would be an answer. So far, that's the only one I have.
450.
69
36
If you can repeat a digit, there are 27. If you can't repeat a digit, there are only 6.
With 123 digits you can make 123 one-digit numbers.
There are 5460 five digit numbers with a digit sum of 22.
5040 different 4 digit numbers can be formed with the digits 123456789. This is assuming that no digits are repeated with each combination.
There are 9 digits that can be the first digit (1-9); for each of these there is 1 digit that can be the second digit (6); for each of these there are 10 digits that can be the third digit (0-9); for each of these there are 10 digits that can be the fourth digit (0-9). → number of numbers is 9 × 1 × 10 × 10 = 900 such numbers.
27 three digit numbers from the digits 3, 5, 7 including repetitions.
To find the even two-digit numbers where the sum of the digits is 5, we need to consider the possible combinations of digits. The digits that sum up to 5 are (1,4) and (2,3). For the numbers to be even, the units digit must be 4, so the possible numbers are 14 and 34. Therefore, there are 2 even two-digit numbers where the sum of the digits is 5.
11
11
Using the digits of 1345678, there are 210 three digit numbers in which no digit is repeated.
N squared would be used to find the square root of a number or numbers. In order to find the number of three digit numbers such that the sum of the square results of any two digits are equal to the third digit the use of the formula (HOE)squared=Hsquared*10000+2HE*100+Esquared is needed.
It has to be something like 11*50=550 (5^2)+(5^2)=50 Therefore, 550 would be an answer. So far, that's the only one I have.
There are 238 - 1 = 237 numbers between 1 and 238. To find the number of digits, we need to consider the range of numbers from 1 to 9 (1-digit numbers), 10 to 99 (2-digit numbers), and 100 to 238 (3-digit numbers). There are 9 one-digit numbers, 90 two-digit numbers, and 139 three-digit numbers between 1 and 238, totaling 9 + 90 + 139 = 238 digits.