How many integers between 1000 and 9999 have distinct digits?
4536
A positive integer. my digits r odd.Im equal to the sum of the cubes of my digits.Im less than 300.who am I?
Okay - the number will either have one, two, or three digits. If it's one digit, a = a^3 and a=1 (this is an answer, but probably not what they're looking for.) If it's two digits, 10a+b = a^3 + b^3. Since 5^3 = 125, a and b must both be less than 5. That leaves 13, 31, 33, or 11. None work. Now - what about three digit numbers? We get the equation: 100a + 10b + c = a^3 + b^3 + c^3 9^3 = 729 7^3 = 343 5^3 = 125 3^3 = 27 1^3 = 1 We need to add up three of these to get the three starting digits. 371 would work (27 + 343 + 1 = 371), but the answer has to be less than 300! So it can't have a 7 or a 9 in it. Since the number only has odd digits, it must be less than 200 - the first digit can't be a 2. That means it would have to have exactly one 5 in it. By trial and error, I found 153 = 125 + 27 + 1. So the answer could be 153 or 1.
How many positive integers less than 1,000,000 have the sum of their digits equal to 17?
we have to take care that a digit doesn't exceed 9. for that, we pre-allot 10 as one of the 6 digits, reallot the remaining 7 using the stars and bars formula, and subtract, ie 22c5 - 6c1*12c5 = 21,582 <-------------
How many positive 3-digit integers are upright?
369