1111111
7
The ones place in odd numbers will be odd (1, 3, 5, 7, 9)
7 tens and 4 ones 5 thousandths = 74.005
2, 3, 7, 8
0.6364
The number that has 7 tens and less ones than tens is 70 minus a certain number of ones. Since there are 7 tens, which is 70, and it needs to have less than 7 ones, the possible values for the ones place are 0 through 6. Therefore, the possible numbers are 70, 71, 72, 73, 74, 75, and 76.
The prime numbers are 3 and 7 but 1 is not considered to be a prime or a composite number
232.107
We can't tell what those numbers are without spaces between them. Eliminate the ones that aren't multiples of 7, like 271, and proceed from there or re-submit the question.
The digits in the ones place of our numbers can be any digit from 0 to 9. This means that for any integer, the ones place will represent the last digit, determining its value in terms of units. For example, in the number 57, the ones place is 7, while in 1234, it is 4. Each number will have a unique digit in the ones place based on its value.
No, binary numbers don't consist of ones and twos, they are ones and zeros.
That's not quantifiable; numbers are infinite. You could choose an outrageously high number, its corresponding negative, and 5 sevens. If you restrict yourself to positive numbers, you could choose 6 ones and 43. The point of it is the 7 numbers need to total 49, whatever they are.