To do these problems think of four boxes, where each box represents a digit.
In the first box, you can only have 1-9, excluding 7 (1, 2, 3, 4, 5, 6, 8, 9), that's 8 numbers. Basically, it's the same as the next three boxes, except zero is also excluded.
In the second, third, and forth boxes, you can have 0-9, excluding 7, which is 9 numbers.
To find the answer, multiply the possibilities by each other.
"Box 1" x "Box 2" x "Box 3" x "Box 4"
8 x 9 x 9 x 9 = 81 x 72 = 5832 possibilities
The answer is 5832 possibilities.
From 1000 to 9999 (inclusive) there are 9999 - 1000 + 1 = 9000 integers. Half of which are even, half of which are odd. So the answer is 9000/2 = 4,500.
9000 integers.
31.how do you solve?
128!
There are no numbers that are divisible by 21 but not by 7 so the "or 21" part of the question can be ignored.There are 166 numbers between 1 and 1000 that are divisible by 6. However, 23 of those are also divisible by 7.So 166 - 23 = 143 numbers.
There are 1,000 positive integers between 1,000 and 9,999, inclusive, that are divisible by nine.
There are 3168 such numbers.
750
The sum of the positive integers from 1,000 to 1,100 inclusive is: 106,050
From 1000 to 9999 (inclusive) there are 9999 - 1000 + 1 = 9000 integers. Half of which are even, half of which are odd. So the answer is 9000/2 = 4,500.
There are 2828 integers between 1000 and 9999.
9000 integers.
There are 2704 of them between 1000 and 10000 inclusive.
648
There are 30 such integers.
333 integers.
All multiples of 3 have digits that add up to a multiple of 3. There are 333 multiples of 3 between 1 and 1000.