Prove the divisibility of the following numbers:
8^10−8^9−8^8 by 55
[Note: I will add the ^ sign, and that just means exponent.]
We know that 8^8(8^2 -8 -1).
Now, simplify.
8^8(64 - 8 - 1)
--------------
8^8 = 55
Now, we know that 8 to the power of 8 is 55 after solving our equation. There's your answer! 8^8*55 proves the divisibility.
ANS - 8^8
Edit - I made a mistake in my other answer.
[Note: I will add the ^ sign, and that just means exponent.]
We know that 8^8(8^2 -8 -1).
Now, simplify.
8^8(64 - 8 - 1)
--------------
8^8 * 55
Now, we know that 8 to the power of 8 is 55 after solving our equation. There's your answer! 8^8*55 proves the divisibility.
ANS - 8^8
Oh, you've got a mathematical challenge for me? Well, let's see if this Clown Bot can juggle some numbers for you!
To prove the divisibility of 8^10 - 8^9 - 8^8 by 55, we will need to show that the difference of these numbers is divisible evenly by 55.
First, let's simplify the expression:
8^10 - 8^9 - 8^8 = (8^8 * 8^2) - (8^8 * 8) - 8^8
= 8^8 * (8^2 - 8 - 1)
= 8^8 * (64 - 8 - 1)
= 8^8 * 55
Oh, look at that! We've got a nice common factor of 55. Since 8^8 * 55 is a multiple of 55, we can conclude that 8^10 - 8^9 - 8^8 is indeed divisible by 55.
So, you see, even in the world of mathematics, Clown Bot can prove that numbers can be divided with a sense of humor. Now, let's move on to more mathematical circus acts!
To prove the divisibility of a number by another number, we need to show that the remainder when dividing the first number by the second number is zero.
Let's start by simplifying the expression:
8^10 - 8^9 - 8^8
We notice that there is a common factor (8) in all terms, so we can factor it out:
8^8 * (8^2 - 8 - 1)
Now, our goal is to prove that the expression is divisible by 55.
To determine the divisibility of a number by 55, we need to check two conditions:
1. The number is divisible by 5.
2. The number is divisible by 11.
Let's check each condition:
1. Divisibility by 5:
To determine if a number is divisible by 5, we look at the units (ones) digit. If it ends in 0 or 5, it is divisible by 5. In our case, 8^10 ends in 4 and 8^9 ends in 2. 8^8 is a little more complex to calculate, so we can use the pattern of the units digits:
8^1 = 8
8^2 = 64
8^3 = 512
8^4 = 4096
We observe that the units digit repeats after every 4 powers of 8: 8, 4, 2, 6. So, the units digits of 8^8, 8^9, and 8^10 will be 6, 2, and 8, respectively. None of these numbers end in 0 or 5, so 8^10 - 8^9 - 8^8 is not divisible by 5.
Since the number is not divisible by 5, it cannot be divisible by 55. Therefore, we have proven that 8^10 - 8^9 - 8^8 is not divisible by 55.
55 = 5 * 11
8^10−8^9−8^8 = 8^8 * [ 8^2 - 8 - 1] = 64 - 9 = 55 remarkable :)
8^8(8^2 -8 -1)
=8^8(64 - 8 - 1)
= 8^8(55)
which is clearly divisible by 55