Question

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

Answer 1

Answer:

Proved

Step-by-step 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.

$Result = \frac{Sum}{3}$

$Result = \frac{3x+3}{3}$

$Result = \frac{3(x+1)}{3}$

$Result = x + 1$

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.

$Result = \frac{5x + 10}{5}$

$Result = \frac{5(x + 2)}{5}$

$Result = x + 2$

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.

$Result = \frac{4x + 6}{4}$

Split

$Result = \frac{4x}{4} + \frac{6}{4}$

$Result = x + 1.5$

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