Why isn't the answer to TMUA-P2Y-PRF-010 shown upfront?

The options are lettered without marking the correct one so you attempt TMUA-P2Y-PRF-010 first — TMUA questions reward spotting the trap before you check the worked solution below.

Do I need a calculator for TMUA-P2Y-PRF-010?

No — calculators are banned in the TMUA. TMUA-P2Y-PRF-010 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-P2Y-PRF-010?

TMUA-P2Y-PRF-010 is Paper P2 · Direct proof (specification code PRF1), level 3. Drill more of the same topic from this page.

TMUA-P2Y-PRF-010 — Paper P2 · Direct proof · Level 3

P2PRF1Level 3~270s target

Spec: PRF1 · No calculator · ~270s

A competition proof claims: if

3 \nmid n

then

n^2 = 3t + 1

for some integer

t

. It argues: either

n = 3k + 1

n = 3k + 2

; in the first case

n^2 = 9k^2 + 6k + 1 = 3(3k^2 + 2k) + 1

; in the second

n^2 = 9k^2 + 12k + 4 = 3(3k^2 + 4k + 1) + 1

; so in both cases the remainder is

1

. Five referees comment. Which referee is right?

Options

The proof fails because $n = 3k + 2$ gives remainder $4$ , not $1$ .
The proof only works for positive $n$ ; negatives are missing.
The proof is valid: the two remainders exhaust $3 \nmid n$ , and both squares genuinely leave remainder $1$ with integer quotients.
The proof assumes $n^2 = 3t + 1$ from the start, so it is circular.

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: modular-cycle. Name the trap first, work the stem by hand, then open the solution below.

Show solution

Every integer is

3k

3k + 1

3k + 2

; excluding

3k

leaves exactly the two cases used. Chain: remainder classification

\Rightarrow

square and regroup

\Rightarrow

3(\cdot) + 1

with integer brackets

3k^2 + 2k

and

3k^2 + 4k + 1

. Trap: mistaking the unreduced

+4

(3k+2)^2

for the final remainder — one more factor of

3

hides inside

3 = 3 \times 1

; negatives are already covered since

k

ranges over all integers.

Keep practising

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Browse more

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Options

The proof fails because

n = 3k + 2

gives remainder

4

, not

1

The proof only works for positive

n

; negatives are missing.

The proof is valid: the two remainders exhaust

3 \nmid n

, and both squares genuinely leave remainder

1

with integer quotients.

The proof assumes

n^2 = 3t + 1

from the start, so it is circular.

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-PRF1-006

A competition proof claims that for every integer n, the number n^3 - n is divisible by 6, arguing as follows: factorise…

Direct proof

Open TMUA-P2-PRF1-006 →

Same trap: modular-cycle

P1M-NumLevel 2

TMUA-P1-MNUM-013

What is the remainder when 2^10 is divided by 7?

Number

Open TMUA-P1-MNUM-013 →

TMUA-P2Y-PRF-010 — Paper P2 · Direct proof · Level 3

Options

Keep practising

Master Direct proof

TMUA-P2Y-PRF-010 — Paper P2 · Direct proof · Level 3

Options

Keep practising

Master Direct proof