I think the answer is 612!
1 (tens digit) which is odd is half of 2 (hundreds digit) and the sum is (6+1+2=9).
423
63
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.
If the number is odd, the last digit is odd. That means the ten's digit must be twice the one's digit. There are only four two-digit numbers where the ten's digit is twice the one's digit: 21, 42, 63, 84. Check which of these satisfy all the clues.
The first 5 digit number is 10,000 and the last is 99,999. Thus there are 90,000 numbers between that range. Half of them are odd & half are even. 45,000 each
423
423
423
3221
by making them half
Eight hundred and twenty-four abc c=1/2a b=1/2c So lowest is b so start with b=1 412 = 7 now make b=2 824 = 14
Truncation is an operation whereby a long decimal representation of a number is stopped after a given number of digits, taking no account of the digit which follows. This is different from rounding where the final digit retained may be increased by 1 if the following digits make a number which is greater than half a unit.
421 and 842
It is 875.It can be 875 or 885. Using the round-half-to-even method, as required by IEEE Standard 754, both would be rounded to 880.
the answer for that is 18 1+8=9 18/2=9
42.21 or 84.42
84.42 or 42.21