Prove that 7^7-7^6 is divisible by 6
7^7-7^6 , I see a common factor of 7^6
= 7^6(7 - 1)
= 6*7^6
any number multiplied by 6 must be divisible by 6
Well, to prove that 7^7 - 7^6 is divisible by 6, we first need to simplify the expression. Let's start by factoring out the common factor of 7^6:
7^7 - 7^6 = 7^6 * (7 - 1)
Now, let's simplify further:
7 - 1 = 6
So, we can rewrite the expression as:
7^7 - 7^6 = 7^6 * 6
Since 6 is a factor of the expression, we can conclude that 7^7 - 7^6 is indeed divisible by 6.
But hey, dividing can be such a chore! Just imagine if math were as easy as telling jokes. It's like the difference between a mathematician and a comedian. One calculates formulas, and the other formulates calctuations!
To prove that 7^7 - 7^6 is divisible by 6, we need to show that the difference is divisible by both 2 and 3.
1. First, let's factor out 7^6 from the expression:
7^7 - 7^6 = 7^6(7 - 1)
2. Simplify the expression:
7^7 - 7^6 = 7^6 * 6
At this point, we see that the expression is a product of 7^6 and 6. Now, we need to show that both factors are divisible by 2 and 3.
3. Divisibility by 2:
7^6 is always even, since raising any number to an even power results in an even number. Thus, 7^6 is divisible by 2.
4. Divisibility by 3:
We can write 6 as 3 * 2. Since 7 = (6 + 1), we have 7^6 = (6 + 1)^6.
Using the binomial theorem to expand (6 + 1)^6, we get:
7^6 = 6C0 * 6^6 * 1^0 + 6C1 * 6^5 * 1^1 + 6C2 * 6^4 * 1^2 + ... + 6C6 * 6^0 * 1^6
Notice that all the terms in the expansion are divisible by 6 except for the last term, which is 1^6. Therefore, 7^6 is divisible by 3.
Since both factors, 7^6 and 6, are divisible by 2 and 3, we can conclude that 7^7 - 7^6 is divisible by 6.
To prove that 7^7 - 7^6 is divisible by 6, we can first simplify the expression.
Starting with 7^7 - 7^6:
= 7^6(7 - 1) (taking out the common factor 7^6)
= 7^6(6)
Now, we have 7^6 multiplied by 6. To determine if this expression is divisible by 6, we can check if 7^6 is divisible by 6 separately.
To check if 7^6 is divisible by 6, we need to determine if it is divisible by both 2 and 3.
First, let's check divisibility by 2:
Any number raised to an even power is always divisible by 2. Since 7 is an odd number, 7^6 is also odd. Therefore, 7^6 is not divisible by 2.
Next, let's check divisibility by 3:
To do this, we need to find the remainder when 7^6 is divided by 3.
Using the property of remainders, we can simplify the expression by reducing the base:
7^6 ≡ (7 mod 3)^6 (where "mod" represents the modulo operator)
7^6 ≡ (1)^6 (since 7 mod 3 = 1)
7^6 ≡ 1 (since any number raised to the power of 0 is equal to 1)
Since 7^6 ≡ 1 (mod 3), we know that 7^6 leaves a remainder of 1 when divided by 3. Therefore, 7^6 is divisible by 3.
Now that we have determined that 7^6 is divisible by 3 but not by 2, we can conclude that 7^6(6) is divisible by both 2 and 3. Therefore, 7^7 - 7^6 is divisible by 6.