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)
Let’s arrange the fist equation
x+y=12
2x+3y=29
Multiply equation 1 by 2
So it will be
2x+2y=24
Subtract eq 1 from eq 2
2x+3y=29
2x+2y=24
—————-
Y =5
Put value of y in eq 1
2x+2y=24
2x +10= 24
2x= 24-10
2x= 14
x=14/2
x=7
S.S= (7,5)
You can do C(6,2) which gives
which is
15!
Answer:
C
Step-by-step explanation:
If I had to make a choice it would be C. Its strange because all numbers can be divided by 4 so they all can make a square. But the closest was C.
