Date | May 2022 | Marks available | 4 | Reference code | 22M.1.SL.TZ2.3 |
Level | Standard Level | Paper | Paper 1 (without calculator) | Time zone | Time zone 2 |
Command term | Prove | Question number | 3 | Adapted from | N/A |
Question
Consider any three consecutive integers, n-1n−1, nn and n+1n+1.
Prove that the sum of these three integers is always divisible by 33.
Prove that the sum of the squares of these three integers is never divisible by 33.
Markscheme
(n-1)+n+(n+1)(n−1)+n+(n+1) (A1)
=3n=3n A1
which is always divisible by 33 AG
[2 marks]
(n-1)2+n2+(n+1)2 (=n2-2n+1+n2+n2+2n+1)(n−1)2+n2+(n+1)2 (=n2−2n+1+n2+n2+2n+1) A1
attempts to expand either (n-1)2(n−1)2 or (n+1)2(n+1)2 (do not accept n2-1n2−1 or n2+1n2+1) (M1)
=3n2+2=3n2+2 A1
demonstrating recognition that 22 is not divisible by 33 or 2323 seen after correct expression divided by 33 R1
3n23n2 is divisible by 33 and so 3n2+23n2+2 is never divisible by 33
OR the first term is divisible by 33, the second is not
OR 3(n2+23)3(n2+23) OR 3n2+23=n2+233n2+23=n2+23
hence the sum of the squares is never divisible by 33 AG
[4 marks]
Examiners report
Most candidates were able to earn full marks in part (a), though some were not able to provide the required reasoning to earn full marks in part (b). In many cases, candidates did not seem to understand the nature of a general deductive proof, and instead substituted different consecutive integers (such as 1, 2,3 ), showing the desired result for these specific values, rather than an algebraic generalization for any three consecutive integers.