Any element of the set of numbers of the form 40*k, where k is an integer, is evenly divisible.
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To be divisible by 36 the number must be divisible by 9 (and 4). To be divisible by 9, the sum of the digits must be divisible by 9. 5 + 5 = 10 → need an extra 8 → A + B = 8.
No, it does not go in evenly. To be divisible by 8 it would need to be divisible by two and 5 being and odd number is not.
Any number ending in 5 or 0 is divisible by 5, just like any number ending in 2, 4, 6, 8 or 0 is divisible by 2.
4000
1200 = 3*5*8*10