Clearly A, B, and C are digits between zero and eight. A takes any value from 1 to 8.* There are thus eight possibilities for the value of A. Each value of A will dictate a value N, from 0 to 7, for (B+C). For a given value of N, B can take any value from 0 to N, and thus can take N + 1 different values, each corresponding to a particular value of C. So there are N + 1 ways of achieving (B+C) = N. The total number of combinations is therefore add up [N+1] between N=0 and N=7 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 = (8 + 1) x 8/2 = 9 x 4 = 36 And here are those 36 solutions: 800,701,710,611,620,602,521,512,530,503,422,431,413,440,404,332,323,341,314,350,305,233,242,224,215,251,206,260,143,134,152,125,116,161,107,170 *If A was zero, we wouldn't really have a three digit number as required.
Assuming you mean permutations of three digits, then the set of numbers that can be made with these digits is: 345 354 435 453 534 543 There are six possible permutations of three numbers.
Three digits.
All three numbers are significant digits, so 3.
21
Assuming that 001, 080, etc are not allowed (that is a leading zero or two is not permitted), the smallest number with exactly three digits is 100. The largest number with exactly three digits is 999. So there are 999 - 100 + 1 = 900 numbers with exactly three digits.
36
there are 899 whole numbers that have three digits.
You can always tell a number is divisible by eight by checking to see if the rightmost three digits are. If they are, then the whole number is. With that in mind then, you can make up all sorts of such numbers by just grabbing some combination that you know is divisible by eight, and tacking additional digits on to the left. For example, we know that 128 is divisible by eight (128 / 8 = 16), so any of these numbers are as well: 89469128 13948128 90173128 The number of digits doesn't matter, so we also know that a ridiculously large number like this one: 25098174509840582437575109283102477568758134024 is divisible by eight, because the last three digits are.
Assuming you mean permutations of three digits, then the set of numbers that can be made with these digits is: 345 354 435 453 534 543 There are six possible permutations of three numbers.
There are 900 three-digit numbers.
Three digits.
depends on your answer
757
100
Yes.
None, because you do not have three even digits. None, because you do not have three even digits. None, because you do not have three even digits. None, because you do not have three even digits.
All three numbers are significant digits, so 3.