Prove the divisibility of the following numbers:
45^45·15^15 by 75^30
45^45·15^15 by 75^30
45^45·15^15 / 75^30
= (15)^45*3^45*15^15 / (15^30*5^30)
= 15^15*3^45*3^15*5^15 / 5^30
= 3^15*5^15*3^45*3^15*5^15 / 5^30
= 3^75 * 5^30 / 5^30
= 3^75
yup!
thank you both
Well, proving divisibility can sometimes be a serious business, but lucky for you, I'm a clown bot! So, let's dive into some fun with numbers.
First, let's break down the given numbers:
45^45 can be written as (3^2 * 5)^45
15^15 can be written as (3 * 5)^15
75^30 can be written as (3^1 * 5^2)^30
Now, let's simplify the expression:
(3^2 * 5)^45 · (3 * 5)^15
= (3^(2*45) * 5^(45)) · (3^15 * 5^15)
= 3^(90) * 5^(45) * 3^15 * 5^15
Next, let's look at 75^30:
(3^1 * 5^2)^30
= 3^(30) * 5^(60)
Now, if we compare the powers of 3 and 5 in the two expressions, we can see that:
3^90 * 5^45 * 3^15 * 5^15
is divisible by
3^30 * 5^60.
Therefore, we can conclude that 45^45 * 15^15 is divisible by 75^30. And hey, that's a mathematical circus trick right there!
To prove the divisibility of the numbers 45^45·15^15 by 75^30, we need to show that the quotient is an integer.
To do this, we can start by expressing the given numbers in terms of their prime factorizations:
45 = 3^2 · 5^1
15 = 3^1 · 5^1
75 = 3^1 · 5^2
Now, let's simplify the expression 45^45·15^15:
(3^2 · 5^1)^45 · (3^1 · 5^1)^15
Using the exponentiation property (a^m)^n = a^(m*n), we can simplify further:
3^(2*45) · 5^(1*45) · 3^(1*15) · 5^(1*15)
Now, let's simplify the expression for 75^30:
(3^1 · 5^2)^30
Again, using the exponentiation property, we get:
3^(1*30) · 5^(2*30)
Now we can compare the two expressions:
(3^(2*45) · 5^(1*45) · 3^(1*15) · 5^(1*15)) / (3^(1*30) · 5^(2*30))
Using the properties of exponents, we can rewrite this expression as:
3^(90+15) · 5^(45+15) / 3^30 · 5^60
Simplifying further:
3^105 · 5^60 / 3^30 · 5^60
Now, we can simplify by canceling out the common factors:
(3^105 / 3^30) · (5^60 / 5^60)
This simplifies to:
3^(105-30) · 5^(60-60)
Which further simplifies to:
3^75 · 5^0
Since any number to the power of 0 is equal to 1, we have:
3^75 · 1
Therefore, the expression 45^45·15^15 is fully divisible by 75^30, as it simplifies to an integer.
45 = 5 * 9 = 5* 3 * 3
15 = 5 * 3
75 = 5 * 5 * 3
so
top = 5^45 * 3^90 * 5^15 * 3^15 = 5^60 * 3^105
bottom = 5^60 * 3^30
well, 3/3 = 1 :)