If you want your answer to be a whole number, it is divisible by 2 and 5.
yes To be divisible by 2 the last digit must be even, ie one of {0, 2, 4, 6, 8}; The last digit of 430 is 0, which is one of these so it is divisible by 2.
There is an infinite number of numbers that are divisible by 430.
hcf(520, 430) = 10 520 = 23 x 5 x 13 430 = 2 x 5 x 43 hcf = 2 x 5 = 10
It is divisible by 2 and 3. It isn't divisible by 5 and 10.
If you want your answer to be a whole number, it is divisible by 2 and 5.
Yes.
2, 5, 6, and 10.
yes To be divisible by 2 the last digit must be even, ie one of {0, 2, 4, 6, 8}; The last digit of 430 is 0, which is one of these so it is divisible by 2.
Ok look, it is divisible by 2, 5, and 10. :-)
The factors of 430 are 2, 5, and 43 or 2 times 5 time 43 is 430.
No. 430 is not evenly divisible by three.
There is an infinite number of numbers that are divisible by 430.
430 = 2 * 5 * 43
Find the sum of the first hundred even natural number divisible by 5?
-- Neither 2 nor 5 is divisible by 18. -- 18 is divisible by 2, but not by 5.
1, 2, 5, 10, 19, 38, 43, 86, 95, 190, 215, 430, 817, 1634, 4085, 8170