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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.
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How many odd integers are there from 1000 through 9999?
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.
How many integers are between 1000 and 10000?
9000 integers.
How many even integers are between 100 and 1000?
To find the number of even integers between 100 and 1000, we first determine the number of even integers between 1 and 1000, which is half of the total integers (since every other integer is even). So, 1000/2 = 500 even integers between 1 and 1000. Next, we subtract the number of even integers between 1 and 100, which is 50 (since every other integer is even in this range as well). Therefore, there are 500 - 50 = 450 even integers between 100 and 1000.
How many integers between 1 and 1000 are the product of two consecutive integers?
31.how do you solve?
How many integers between 100 and 1000 are multiples of 7?
128!
