11,111
2457
For a number to have 5 as a factor, its ones place must be either 0 or 5. This is because any number that ends in 0 or 5 is divisible by 5. Therefore, the digits that can be in the ones place of a number with 5 as a factor are 0 and 5.
7777 + 777*4 + 7*22 + 77 - 7 + 2 = 11111
the 5 as it is 5 thousandths
7 in the ones place
2457
For a number to have 5 as a factor, its ones place must be either 0 or 5. This is because any number that ends in 0 or 5 is divisible by 5. Therefore, the digits that can be in the ones place of a number with 5 as a factor are 0 and 5.
If the original number was even, the ones place will be zero; otherwise it will be 5.
0 or 5
7777 + 777*4 + 7*22 + 77 - 7 + 2 = 11111
If 5 is a factor of a number, it means the number is divisible by 5. For the ones digit of the number to be 5, the number must end in 5. Since there are 10 possible digits (0-9), and only one of those digits is 5, the probability of the ones digit being 5 is 1 out of 10, or 1/10.
2005The ones digit is 5.If the ones digit is 5 or 0 then it is divisible by5.NOT a prime.
Factor them both. 1 x 5 = 5 1 x 7 = 7 Discard the common factor (1) and combine the remaining ones 5 x 7 = 35 35 is the lowest (least) number that both 5 and 7 will go into evenly.
the 5 as it is 5 thousandths
When a number is a multiple of 5, the possible values of the ones digit are zero and five.
7 in the ones place
In the number 235, there is one "1" in the tens place, and the digit in the ones place is 5. Therefore, if you are asking how many individual ones are in the number 235, the answer is 5.