If the first digit is 9, you have 9 options (0-8) for the second digit.
If the first digit is 8, you have 8 options (0-7) for the second digit.
Etc.
This leaves you with the arithmetic series: 0 + 1 + 2 + 3 + ... + 9.
4082 Since the tens digit is 2 times the thousand digit, it must be an even digit. So it can be 8, 6, 4, or 2. But, the thousands digit is 4 greater than the hundred digit. So that the hundred digit must be 0, the thousands digit must be 4, the hundreds digit must be 8, and the ones digit must be 2.
The "6" is in the ones place.
No. The tens unit is independent of the ones unit
875 rounded to nearest ten is 880 8+8+0 is 16
start with the 100's place 0 = 0+0 1 = 1+0 = 0+1 2 = 1+1=2+0=0+2 3 = 3+0 = 2+1 = 1+2 = 0+3 ... with the 100's digit, the number of such numbers is one greater than that number 0 : 1 1 : 2 ... therefore there are 1+2+3+4+5+6+7+8+9+10 = 5*11 = 55 such numbers unless you don't count 000 as a three digit number, then there are 54 such numbers.
16
23
Probably 73.
It is 0.45
There are 45 of them.
9218 9425
452, 894. None other that I can think of
89 is a prime number between 70 and 100 that has the ones digit one greater than the tens digit.
The number must be 42.
4
It is 9 times greater.
The answer depends on what the tens digit is greater than, and what the ones digit does then.