2457
7777 + 777*4 + 7*22 + 77 - 7 + 2 = 11111
the 5 as it is 5 thousandths
7 in the ones place
To find if a number is divisible by 5, the digit in the ones place has to be 0 or 5.
2457
If the original number was even, the ones place will be zero; otherwise it will be 5.
0 or 5
2005The ones digit is 5.If the ones digit is 5 or 0 then it is divisible by5.NOT a prime.
7777 + 777*4 + 7*22 + 77 - 7 + 2 = 11111
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.
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.
7 in the ones place
In the number 435, there are 4 tens and 3 ones. The digit 4 represents 4 tens, while the digits 3 and 5 represent 3 ones each. This can be broken down as (4 x 10) + (3 x 1) = 40 + 3 = 43.
Only one positive prime number has a 5 in the ones digit. That prime number is 5. All other numbers with a 5 in the ones digit are composite because they will be divisible by 5.