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.
875 rounded to nearest ten is 880 8+8+0 is 16
No. The tens unit is independent of the ones unit
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
The prime number between 20 and 30 whose ones digit is 7 greater than its tens digit is 23. The tens digit is 2, and the ones digit is 3, which is 7 greater than 2. Additionally, 23 is a prime number because it is only divisible by 1 and itself, with no other factors.
The number must be 42.
89 is a prime number between 70 and 100 that has the ones digit one greater than the tens digit.
4
It is 9 times greater.