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Prove the sum of three consecutive is divisible by three, that the sum of 5 consecutive integers is divisible by 5, but the sum of four consecutive integers is not divisible by 4

User Dtar
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1 Answer

8 votes

Answer:

Proved

Explanation:

Solving (a):

Let the numbers be:
x, x + 1, x + 2.

Their sum is:


Sum = x + x + 1 + x + 2

Collect Like Terms


Sum = x + x + x + 1 + 2


Sum = 3x + 3

Divide sum by 3.


Result = (Sum)/(3)


Result = (3x+3)/(3)


Result = (3(x+1))/(3)


Result = x + 1

Hence, this is true because there is no fractional part after the division

Solving (b):

Let the numbers be:
x, x + 1, x + 2,x+3,x+4

Their sum is:


Sum = x + x + 1 + x + 2+x + 3 + x + 4

Collect Like Terms


Sum = x + x + x + x + x+1 + 2 + 3 + 4


Sum = 5x+10

Divide sum by 5.


Result = (5x + 10)/(5)


Result = (5(x + 2))/(5)


Result = x + 2

Hence, this is true because there is no fractional part after the division

Solving (c):

Let the numbers be:
x, x + 1, x + 2,x+3

Their sum is:


Sum = x + x + 1 + x + 2+x + 3

Collect Like Terms


Sum = x + x + x + x + 1 + 2 + 3


Sum = 4x+6

Divide sum by 4.


Result = (4x + 6)/(4)

Split


Result = (4x)/(4) + (6)/(4)


Result = x + 1.5

The 1.5 means that the sum can not be divisible by 4

User WalterBeiter
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