Show that one of the three consecutive odd integer
is a multiple of 3.
Answers
Answer:
Hey mate
Here is your answer
Since it is an odd number
So
1st number be 2n+1
Being consecutive odd
2nd number will be 2n+1+2=2n+3
And
3rd number 2n+5
2n is always an even number and condition will revolve around this only giving rise to 2 cases.
Case 1 : if 2n is divisible by 3
then 2n + 3 will always be divisible by 3.
Case 2 : if 2n is not divisible by 3
Then remainder is either 1 or 2
Case 2 (a). Remainder 1 implies 2n-1 is divisible by 3.
If multiple of 3 is added to 2n-1 that will also be divisible by 3
2n-1 +3 + 3
=2n-1 +2×3
=2n+5
So 2n+5 is divisible by 3
Case 2 (b) Remainder 2 implies 2n-2 is divisible by 3.
If multiple of 3 is added to 2n-2 that will also be divisible by 3
2n-2 +3
=2n-2 +1×3
=2n+1
So 2n+1 is divisible by 3
So from all the above case it os verified that out of 3 consecutive odd number one will always be divisible by 3.
Hope it helps...
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X+x+3+x+6 = 0
3x +9 = 0
X= -9/3
X=-3
X. =. -3