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28 is divisible by...
1, because everything is.
2, because it's even.
4, because the last two digits are divisible by 4.
Divide them into 28.
All Factors of 28:
1, 2, 4, 7, 14, 28
Wiki User
∙ 6y ago
Wiki User
∙ 6y agoDivisibility by 28 requires divisiblity by 4 AND divisibility by 7.
- Divisibility by 4 can be established by looking at the integer formed by the last two digits (the Tens and Units) digit. Divide this number by 2. If the quotient is even (end in 0, 2, 4, 6, 8), then the given number is divisible by 4.
- Divisibility by 7 is not easy to establish. The following is the only method that I know of.
- Partition the number by separating the last digit. Call the leading part L and the rest R.
- Calculate L - 2R.
- Repeat if necessary until you get a small number.
- If the last number is a multiple of 7 (incl 0), then the number is divisible by 7.
Example:
Given number 15848.
The number formed by the last two digits is 48. Half of 84 is 24 which is even. So the number is divisible by 4.
Partition # 1: L = 1584 and R = 8, L - 2R = 1568
Partition # 2: L = 156 and R = 8, L - 2R = 140
Partition # 3: L = 14 and R = 0, L - 2R = 14
Partition # 4: L = 1 and R = 4, L - 2R = -7 which is a multiple of 7.
You may have twigged after partition 2, result = 140 was a multiple of 7!
So, the number is a multiple of 4 and of 7 and so it is a multiple of 28. In fact, 15848 = 28*566.
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