Prove the divisibility of the following numbers:
810−89−88 by 55
Seems like Much Ado About Nothing,
any number that ends in 0 or 5 is divisible by 5
Which of your numbers fits that condition?
To prove the divisibility of a number by another number, we need to check whether the remainder when the first number is divided by the second number is zero.
Let's divide 810−89−88 by 55:
810−89−88 ÷ 55
To make the division process easier, let's simplify the expression:
810 - 89 - 88 = 810 - (89 + 88) = 810 - 177 = 633
Now, let's divide 633 by 55:
633 ÷ 55 = 11 with a remainder of 8
Since the remainder is not zero, we can conclude that 633 is not divisible by 55. Therefore, we can also conclude that 810−89−88 is not divisible by 55.