Question 3
The product of three consecutive positive integers is divisible by 6: Is this statement true or false? Justify your answer.
Question 3
The product of three consecutive positive integers is divisible by 6: Is this statement true or false? Justify your answer.
The statement is true.
When 3 is the divisor, any positive integer can be expressed as 3q+r, where q is any integer ,and 0≤r<3
The possible values of r are 0,1 and 2.
Thus, any integer can be expressed as 3q, 3q+1 or 3q + 2
Case 1:
Consider the first integer to be 3q
The consecutive integers will be 3q + 1 and 3q + 2.
3q is divisible by 3.
When q is even, then 3q is even, which means it is divisible by 2.
When q is odd, 3q is odd. 3q + 1 is even, which means it is divisible by 2.
Thus, the product of 3q, 3q + 1 and 3q + 2 is divisible by both 2 and 3. Thus, the product of the three consecutive integers is divisible by 6
Case 2:
Consider the first integer to be 3q + 1
The consecutive integers will be 3q + 2 and 3q + 3.
3q + 3 is divisible by 3.
When q is even, then 3q is even. This implies that 3q + 2 is even and is divisible by 2.
When q is odd, 3q is odd. 3q + 1 is even, which means it is divisible by 2.
Thus, the product of 3q + 1, 3q + 2 and 3q + 3 is divisible by both 2 and 3. Thus, the product of the three consecutive integers is divisible by 6
Case 3:
Consider the first integer to be 3q + 2
The consecutive integers will be 3q + 3 and 3q + 4.
3q + 3 is divisible by 3.
When q is even, then 3q is even. This implies that 3q + 4 is even and is divisible by 2.
When q is odd, 3q is odd. 3q + 3 is even, which means it is divisible by 2.
Thus, the product of 3q + 2, 3q + 3 and 3q + 4 is divisible by both 2 and 3. Thus, the product of the three consecutive integers is divisible by 6
Hence, the product of three consecutive integers is divisible by 6.
The statement is true.
When 3 is the divisor, any positive integer can be expressed as 3q+r, where q is any integer ,and 0≤r<3
The possible values of r are 0,1 and 2.
Thus, any integer can be expressed as 3q, 3q+1 or 3q + 2
Case 1:
Consider the first integer to be 3q
The consecutive integers will be 3q + 1 and 3q + 2.
3q is divisible by 3.
When q is even, then 3q is even, which means it is divisible by 2.
When q is odd, 3q is odd. 3q + 1 is even, which means it is divisible by 2.
Thus, the product of 3q, 3q + 1 and 3q + 2 is divisible by both 2 and 3. Thus, the product of the three consecutive integers is divisible by 6
Case 2:
Consider the first integer to be 3q + 1
The consecutive integers will be 3q + 2 and 3q + 3.
3q + 3 is divisible by 3.
When q is even, then 3q is even. This implies that 3q + 2 is even and is divisible by 2.
When q is odd, 3q is odd. 3q + 1 is even, which means it is divisible by 2.
Thus, the product of 3q + 1, 3q + 2 and 3q + 3 is divisible by both 2 and 3. Thus, the product of the three consecutive integers is divisible by 6
Case 3:
Consider the first integer to be 3q + 2
The consecutive integers will be 3q + 3 and 3q + 4.
3q + 3 is divisible by 3.
When q is even, then 3q is even. This implies that 3q + 4 is even and is divisible by 2.
When q is odd, 3q is odd. 3q + 3 is even, which means it is divisible by 2.
Thus, the product of 3q + 2, 3q + 3 and 3q + 4 is divisible by both 2 and 3. Thus, the product of the three consecutive integers is divisible by 6
Hence, the product of three consecutive integers is divisible by 6.
