What is the smallest multiple of 28 that has all the same number?

Question

Fun little problem for ya.

Answer 1

The answer is: 444,444.
28=4*7. The smallest number which has the same numbers, should be made of some "4" to be dividable to 4. For example: 44, 444 and etc. if you test them, you can find the smallest one which is dividable to 7 and so dividable to 28 is 444,444.

Answer 2

Such a number would be a multiple of a number with all ones. Because "all ones" is an odd number, we cannot do much with the factors two in 28, so the final number will have all fours or all eights. So we need an "all ones" number that is the multiple of 7.

A multiple of 7 has the property that

(last digit) +
3*(second last digit) +
2*(third last digit) -
(fourth last digit) -
3*(fifth last digit) -
2*(sixth last digit) -
(seventh last digit) + ... [pattern repeats]

is a multiply of seven. If all digits are one, we get
1 + 3 + 2 - 1 - 3 - 2 + 1 + 3 + 2 - ... must be a multiple of 7, and it is easy to see that that happens first with
1 + 3 + 2 - 1 - 3 - 2 = (1+3+2)-(1+3+2) = 6-6 = 0. The solution therefore has four digits, and you get

444444
888888

The third smallest solution is
444444444444

Answer 3

It would be some multiple of 11, 111, 1111, 11111, 111111, etc.

Let's figure the prime factors of these numbers. We need to find one that has 4 or 7 as a factor.
11 --> 11
111 --> 3 * 37
1,111--> 11 * 101
11,111 --> 41 * 271
111,111 --> 3 * 7 * 11 * 13 * 37

Finally, this last one has one of the factors of 28 (namely 7) that will leave a single digit factor remaining (namely 4), so if we multiply by 4, we get:
= 2 * 2 * 3 * 7 * 11 * 13 * 37
= 28 * 15,873
= 444,444

Answer 4

444,444

Answer 5

444,444