Let n be a natural number greater than or equal to 3 such that n-3 is a multiple of 4. Show that n squared 6n 5 is a multiple of 16.
We know that n - 3 is a multiple of 4. Let's express n - 3 as 4k, where k is a positive integer.
Adding 3 to both sides, we have n = 4k + 3.
Now, let's substitute this expression for n in the expression n² + 6n + 5:
n² + 6n + 5 = (4k + 3)² + 6(4k + 3) + 5.
Expanding and simplifying, we have:
n² + 6n + 5 = 16k² + 24k + 9 + 24k + 18 + 5
= 16k² + 48k + 32
= 16(k² + 3k + 2).
Since k² + 3k + 2 is an integer, let's call it m.
Therefore, n² + 6n + 5 is equal to 16m, showing that it is a multiple of 16.
Thus, we have shown that if n - 3 is a multiple of 4, then n² + 6n + 5 is a multiple of 16.
Let's break down the problem step by step:
Step 1: Assume n - 3 is a multiple of 4.
Let's assume that n - 3 = 4k, where k is an integer.
Step 2: Solve for n.
Adding 3 to both sides of the equation, we get:
n = 4k + 3
Step 3: Substitute n into the expression n^2 + 6n + 5.
Substituting the value of n from Step 2 into the expression:
n^2 + 6n + 5 = (4k + 3)^2 + 6(4k + 3) + 5
Step 4: Expand and simplify the expression.
Expanding and simplifying the expression:
(4k + 3)^2 + 6(4k + 3) + 5 = 16k^2 + 24k + 9 + 24k + 18 + 5
= 16k^2 + 48k + 32
= 16(k^2 + 3k + 2)
Step 5: Show that the expression is a multiple of 16.
Since k is an integer, k^2 + 3k + 2 is also an integer. Let's call this integer m.
Substituting m into the expression:
16(k^2 + 3k + 2) = 16m
Therefore, n^2 + 6n + 5 is a multiple of 16, as it can be expressed as 16m for some integer m.
To summarize:
Assuming n - 3 is a multiple of 4, we showed that n^2 + 6n + 5 is a multiple of 16.
To show that n^2 + 6n + 5 is a multiple of 16, we need to prove that it is divisible by 16 without any remainder.
Given that n-3 is a multiple of 4, we can write n-3 as 4k, where k is an integer.
Adding 3 to both sides of the equation, we get n = 4k + 3.
Substituting this value of n into the expression n^2 + 6n + 5, we have:
(4k + 3)^2 + 6(4k + 3) + 5
Expanding this, we get:
16k^2 + 24k + 9 + 24k + 18 + 5
Combining like terms, we have:
16k^2 + 48k + 32
Factoring out 16 from each term, we get:
16(k^2 + 3k + 2)
Since k is an integer, k^2 + 3k + 2 is also an integer.
Therefore, n^2 + 6n + 5 is divisible by 16, as it can be expressed as 16 multiplied by an integer.
Hence, we have shown that if n is a natural number greater than or equal to 3 such that n-3 is a multiple of 4, then n^2 + 6n + 5 is a multiple of 16.