Proving with odd, even, and consecutive

Mathematics · Mathematical Thinking · ages 15-16

Name ______________________ Date ____________

Two consecutive integers are 14 and 15. What is their product?
Answer: ______________
Which form describes ANY even number?
- 2n
- 2n + 1
- n + 2
- 2n - 1
Which form describes ANY even number?
- 2n + 1
- 2n
- n + 2
Sam says testing 1000 examples proves a claim for all integers. Is Sam right?
Circle one: True False
Two consecutive integers add to 2n + 1 with n = 20. What is their total?
Answer: ______________
Which pair MUST have an even product?
- 9 and 14
- 3 and 5
- 5 and 7
Why is n(n + 1) always even?
- n is always even
- n + 1 is always odd
- One of n, n+1 is even
- Products are always even
Maya says in every consecutive pair exactly one number is even. Is Maya right?
Circle one: True False
Three consecutive integers start at 20. What is their sum?
Answer: ______________
With n = 12, what is n(n + 1)?
Answer: ______________

Answer key

For grown-ups. Fold this page away before handing over the rest.

210 · 14 times 15 = 210.
2n · 2n is a multiple of 2 for every integer n. The 2n + 1 option is always odd.
2n · 2n is a multiple of 2 for every integer n.
False · No test count covers infinity; only algebra over n settles every case.
41 · 2 times 20 plus 1 = 41.
9 and 14 · 14 is even so the product is even, while each other pair multiplies two odds.
One of n, n+1 is even · Neighbours alternate parity, so the pair always contains a multiple of 2. The products claim fails, since 3 times 5 is 15.
True · Parity alternates, so neighbours always split odd and even.
63 · 20 plus 21 plus 22 is 63, a multiple of three.
156 · 12 times 13 = 156.