No, because then you would have a remainder of 2. The divisibility rule for 5 is that it needs to end in a zero or a five.
Yes; it is 87.
87 is divisible by 1, 3, 29, and 87. 100 is divisible by 1, 2, 4, 5, 10, 20, 25, 50, and 100, therefore the only number divisible by 87 and 100 is 1.
the numbers divisible by 87 are 1,3,29, and 87
numbers divisible to 87 = 3, 29, 87
No. 87 is not evenly divisible by nine.
85 is not a prime number as it is divisible by 5. Anything that ends with 5 is divisible by 5. The nearest prime number to 85 is 83 or 87.
1, 3, 5, 15, 29, 87, 145, 435
87/2=43.5
The factors of 87 are: 1, 3, 29, 87
no. it is divisible by 3 (87/3=29)
Take any multiple of 87, and you get a number that is divisible by 87 - for example:87 x 0 87 x 1 87 x 2 87 x 3 ... 87 x (-1) 87 x (-2) ...
87 is not divisible by 4. It must be 84 or 88.