450.
(Note: "x" below signifies that this digit is temporarily being ignored)
This is true for:
10x1
11x2
12x3
13x4
14x5
15x6
16x7
17x8
18x9
So for the first eligible 1000 numbers there are 9 (when we disregard the third digit).
For the next 1000 (that is 2000 to 2999) there will be 8 (it is one less because the first digit is 2 and not 1)
This pattern will continue all the way through the 4 digit numbers, so that our answer is:
9+8+7+6+5+4+3+2+1 = 45.
However as "x" above (that we temporarily ignored) can be any digit we have to multiply this by 10 (as there are 10 digits). Therefore the answer will be 450!
21.
Two (or four) digits added together cannot equal 42. Two-digit numbers multiplied together cannot equal 82.
21
Since the first integer of the 3-digit numbers must be 2, the sum of the second and the third digits would be 13.Since the largest digit is 9, the 3-digit numbers could be 249, 294, 258, 285, 267, and 276.Thus, there are 6 integers between 200 and 300 whose sum of their digits is 15.
No two numbers that can be completely written down with digits can be added, subtracted, multiplied, or divided to equal PI. If they could be divided to equal PI, then PI would be a rational number. But it isn't.
21.
Two (or four) digits added together cannot equal 42. Two-digit numbers multiplied together cannot equal 82.
21
There are 15 of them.
1348.
There are no 3-digit whole numbers whose digits sum to 3. The smallest 3-digit number is 100, and the largest is 999, but in neither case is the sum of the digits equal to 3.
002+002+001=005
There are no three-digit numbers that equal 17. In fact, there are no numbers with more or less than two digits that equal 17. In fact, in the whole infinite supply of numbers, there is only one single number that equals 17. That number is . . . . . . . 17 .
I'm sure there are more than 2 prime numbers that are 400 digits long.
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.
No digit that is 5 places long is equal to 6 times 2 other than 12.000
This is only possible if one of the digits is equal to zero. There are 90 3-digit numbers with a zero in the 10's place, and 90 3-digit numbers with a zero in the 1's place - and 9 numbers that have both a zero in the 10's place and a zero in the 1's place; these would be counted double if you just add the first two. So, you get: 90 + 90 - 9 such numbers.