Hello :
all n in N ; n(n+1)(n+2) = 3a a in N or : <span>≡ 0 (mod 3)
1 ) n </span><span>≡ 0 ( mod 3)...(1)
n+1 </span>≡ 1 ( mod 3)...(2)
n+2 ≡ 2 ( mod 3)...(3)
by (1), (2), (3) : n(n+1)(n+2) ≡ 0×1×2 ( mod 3) : ≡ 0 (mod 3)
2) n ≡ 1 ( mod 3)...(1)
n+1 ≡ 2 ( mod 3)...(2)
n+2 ≡ 3 ( mod 3)...(3)
by (1), (2), (3) : n(n+1)(n+2) ≡ 1×2 × 3 ( mod 3) : ≡ 0 (mod 3) , 6≡ 0 (mod)
3) n ≡ 2 ( mod 3)...(1)
n+1 ≡ 3 ( mod 3)...(2)
n+2 ≡ 4 ( mod 3)...(3)
by (1), (2), (3) : n(n+1)(n+2) ≡ 2×3 × 4 ( mod 3) : ≡ 0 (mod 3) , 24≡ 0 (mod3)
Answer:
The solution for y = -1.
Step-by-step explanation:
Given:
....(1)
....(2)
So, to solve for y, first we solve for x in equation (2):
⇒
Dividing both sides by 2,
⇒
Now, substituting the value of x in equation (1):
⇒
⇒
⇒
⇒
⇒
Dividing both sides by 11,
⇒
Therefore, the solution for y = -1.
When you get to the point where the width is larger than the length, you can stop because you have exhausted all the possibilities. The greatest area will be the rectangle that is closer in to a square than any of the others.
Answer is 8 hope this helps
