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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-1n1, nn and n+1n+1.

Prove that the sum of these three integers is always divisible by 33.

[2]
a.

Prove that the sum of the squares of these three integers is never divisible by 33.

[4]
b.

Markscheme

(n-1)+n+(n+1)(n1)+n+(n+1)          (A1)

=3n=3n           A1

which is always divisible by 33           AG

 

[2 marks]

a.

(n-1)2+n2+(n+1)2   (=n2-2n+1+n2+n2+2n+1)(n1)2+n2+(n+1)2   (=n22n+1+n2+n2+2n+1)           A1

attempts to expand either (n-1)2(n1)2 or (n+1)2(n+1)2    (do not accept n2-1n21 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]

b.

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.

a.
[N/A]
b.

Syllabus sections

Topic 1—Number and algebra » SL 1.6—Simple proof
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Topic 1—Number and algebra

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