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.
To determine how many digit numbers can be formed using the digits 2, 3, 5, 7, and 8, we need to consider the number of digits in the numbers we are forming. For a 1-digit number, we can use any of the 5 digits. For a 2-digit number, we can choose 2 out of the 5 digits and arrange them, giving us (5 \times 4) combinations. We can continue this for 3-digit, 4-digit, and 5-digit numbers, which will yield (5), (20), (60), and (120) respectively. Therefore, the total number of digit numbers is (5 + 20 + 60 + 120 = 205).
There are 5460 five digit numbers with a digit sum of 22.
To find how many two-digit numbers have digits whose sum is a perfect square, we first note that the two-digit numbers range from 10 to 99. The possible sums of the digits (tens digit (a) and units digit (b)) can range from 1 (1+0) to 18 (9+9). The perfect squares within this range are 1, 4, 9, and 16. Analyzing each case, we find the valid combinations for each perfect square, leading to a total of 36 two-digit numbers whose digits sum to a perfect square.
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.
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.
Using the digits of 1345678, there are 210 three digit numbers in which no digit is repeated.
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