Why isn't the answer to TMUA-P2-PRF1-006 shown upfront?

The options are lettered without marking the correct one so you attempt TMUA-P2-PRF1-006 first — TMUA questions reward spotting the trap before you check the worked solution below.

Do I need a calculator for TMUA-P2-PRF1-006?

No — calculators are banned in the TMUA. TMUA-P2-PRF1-006 is set for a no-calculator target of about 270 seconds; the worked solution shows the fastest by-hand route.

Which TMUA specification point is TMUA-P2-PRF1-006?

TMUA-P2-PRF1-006 is Paper P2 · Direct proof (specification code PRF1), level 3. Drill more of the same topic from this page.

TMUA-P2-PRF1-006 — Paper P2 · Direct proof · Level 3

P2PRF1Level 3~270s target

Spec: PRF1 · No calculator · ~270s

A competition proof claims that for every integer

n

, the number

n^3 - n

is divisible by

6

, arguing as follows: factorise

n^3 - n = (n-1)n(n+1)

as three consecutive integers, note that at least one is a multiple of

3

and at least one is even, then conclude divisibility by

6

since

2

and

3

are coprime. Five referees comment on whether this chain is complete and where a careless contestant would slip. Which referee is right?

Options

The proof shows divisibility by $5$ , not $6$ .
The proof only works for positive $n$ .
The proof fails because three consecutive integers need not contain a multiple of $3$ .
The proof fails because an even number and a multiple of $3$ might be the same factor.
The proof is valid: the factors supply coprime divisors $2$ and $3$ , whose product divides the whole.

Letters only — the answer key is hidden. Work it out, then check the solution below.

Hint (check trope)

No written hint yet — check the trope: assumes-what-proves. Name the trap first, work the stem by hand, then open the solution below.

Show solution

Among any three consecutive integers one is

0 \pmod 3

and (a possibly different one) is

0 \pmod 2

; coprimality of

2

3

gives

6 \mid (n-1)n(n+1)

. Overlap is harmless: if one factor supplies both,

6

still divides it. Trap: demanding distinct factors, or doubting consecutive coverage — both misread the modular pigeonhole.

Keep practising

Same topic · same level
P2PRF1Level 3
TMUA-P2-PRF1-011
A student proves directly that n^2+n is even for every integer n. Which argument is valid?
Direct proof
Open TMUA-P2-PRF1-011 →
Same trap: assumes-what-proves
P2ERR1Level 2
TMUA-P2-ERR1-008
A “proof” that the sum of two odds is odd reads: (1) Assume odd m, n with m + n odd. (2) Write m = 2j+1, n = 2k+1. (3) T…
Identifying errors in proofs
Open TMUA-P2-ERR1-008 →
More in this topic
P2PRF1Level 3
TMUA-P2-PRF1-012
Which of the following is a valid direct proof that if 0 < x < y then x^2 < y^2?
Direct proof
Open TMUA-P2-PRF1-012 →

Browse more

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Options

The proof shows divisibility by

5

, not

6

The proof only works for positive

n

The proof fails because three consecutive integers need not contain a multiple of

3

The proof fails because an even number and a multiple of

3

might be the same factor.

The proof is valid: the factors supply coprime divisors

2

and

3

, whose product divides the whole.

Letters only — the answer key is hidden. Work it out, then check the solution below.

Keep practising

Same topic · same level

P2PRF1Level 3

TMUA-P2-PRF1-011

A student proves directly that n^2+n is even for every integer n. Which argument is valid?

Direct proof

Open TMUA-P2-PRF1-011 →

Same trap: assumes-what-proves

P2ERR1Level 2

TMUA-P2-ERR1-008

A “proof” that the sum of two odds is odd reads: (1) Assume odd m, n with m + n odd. (2) Write m = 2j+1, n = 2k+1. (3) T…

Identifying errors in proofs

Open TMUA-P2-ERR1-008 →

TMUA-P2-PRF1-006 — Paper P2 · Direct proof · Level 3

Options

Keep practising

Master Direct proof

TMUA-P2-PRF1-006 — Paper P2 · Direct proof · Level 3

Options

Keep practising

Master Direct proof