1,2,4,7,14,28
28=2 squared times 7
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To be divisible by 5, the last digit must be 0 or 5. The last digit of 28 is neither 0 nor 5, so it is not divisible by 5.
Yes - 252/9 = 28
56 is divisible by 1, 2, 4, 7, 8, 14, 28, and 56.
No - every other multiple of 4 is 4 12 20 28 etc are divisible by 4 but not by 8
Whole numbers are divisible by 4 if the number formed by the last two individual digits is evenly divisible by 4. For example, the number formed by the last two digits of the number 2628 is 28, which is evenly divisible by 4 so the number 2628 is evenly divisible by 4.