Level 1 — RecognitionNumber Theory (Intermediate)

Number Theory (Intermediate)

⏱ 20 minutes30 marksprintable — key stays hidden on paper

Time limit: 20 minutes Total marks: 30

Section A — Multiple Choice (1 mark each)

Choose the single best answer.

Q1. Which of the following numbers is divisible by $11$ ? (a) $12321$ (b) $90728$ (c) $45678$ (d) $10001$

Q2. The decimal expansion of $\dfrac{7}{20}$ is: (a) terminating (b) purely repeating (c) eventually repeating (non-terminating) (d) irrational

Q3. Which statement correctly places the number systems? (a) $\mathbb{R} \subset \mathbb{Q} \subset \mathbb{Z} \subset \mathbb{N}$ (b) $\mathbb{N} \subset \mathbb{Z} \subset \mathbb{Q} \subset \mathbb{R}$ (c) $\mathbb{Z} \subset \mathbb{N} \subset \mathbb{Q} \subset \mathbb{R}$ (d) $\mathbb{Q} \subset \mathbb{N} \subset \mathbb{Z} \subset \mathbb{R}$

Q4. The value of $|-7| + |3| - |-2|$ is: (a) $8$ (b) $12$ (c) $-6$ (d) $2$

Q5. $\gcd(48, 36)$ equals: (a) $6$ (b) $4$ (c) $12$ (d) $144$

Q6. Which of these is a solution to $x \equiv 2 \pmod 5$ ? (a) $12$ (b) $15$ (c) $22$ (d) $8$

Q7. Fermat's Little Theorem states that for a prime $p$ and integer $a$ with $p \nmid a$ : (a) $a^{p} \equiv 1 \pmod p$ (b) $a^{p-1} \equiv 1 \pmod p$ (c) $a^{p} \equiv 0 \pmod p$ (d) $a^{p-1} \equiv a \pmod p$

Q8. A number is divisible by $6$ if and only if it is divisible by: (a) $2$ and $4$ (b) $3$ and $9$ (c) $2$ and $3$ (d) $6$ only

Q9. By the Extended Euclidean Algorithm, integers $x,y$ with $ax+by=\gcd(a,b)$ are guaranteed by: (a) Fermat's theorem (b) Bézout's identity (c) the CRT (d) the division algorithm alone

Q10. Which number is irrational? (a) $\sqrt{16}$ (b) $0.\overline{3}$ (c) $\sqrt{2}$ (d) $\dfrac{22}{7}$

Section B — Matching (5 marks)

Q11. Match each number/expression in Column A with its correct property in Column B. Write the pairing (e.g. i–c). (1 mark each)

Column A	Column B
(i) $\sqrt{2}$	(a) terminating decimal
(ii) $\dfrac{1}{8}$	(b) divisible by 3 and 9
(iii) $17 \bmod 4$	(c) irrational
(iv) $\dfrac{2}{3}$	(d) equals 1
(v) $99$	(e) repeating decimal

Section C — True/False WITH Justification (2 marks each: 1 mark verdict, 1 mark justification)

Q12. True or False: The number $123456$ is divisible by $3$ . Justify.

Q13. True or False: Every integer is a rational number. Justify.

Q14. True or False: $\gcd(a,b)=1$ means $a$ and $b$ are both prime. Justify.

Q15. True or False: $7 \equiv -3 \pmod{10}$ . Justify.

Q16. True or False: $\pi$ is a rational number because $\pi \approx \dfrac{22}{7}$ . Justify.

Q17. True or False: The number $73$ is divisible by $7$ (using the divisibility rule for 7). Justify.

Q18. True or False: If $\gcd(m,n)=1$ , the Chinese Remainder Theorem guarantees a unique solution modulo $mn$ to a pair of congruences $x\equiv a\pmod m$ , $x\equiv b\pmod n$ . Justify.

Answer keyMark scheme & solutions

Section A (1 mark each)

Q1. (a) $12321$ . Rule for 11: alternating sum of digits $\equiv 0 \pmod{11}$ . $1-2+3-2+1 = 1$ … recompute right-to-left: $1-2+3-2+1=1$ . Check others properly: For $12321$ : alternating sum $= (1+3+1)-(2+2)=5-4=1$ — not divisible. Correct answer is (d) $10001$ : $(1+0+1)-(0+0)=2$ no. Re-evaluate cleanly:

$10001$ : digits $1,0,0,0,1$ ; alt sum $=1-0+0-0+1=2$ → no.
$90728$ : $9-0+7-2+8=22\equiv0\pmod{11}$ → divisible. Answer: (b) $90728$ . (1 mark) — alternating sum $=22$ , a multiple of 11.

Q2. (a) terminating. $\dfrac{7}{20}=\dfrac{7}{2^2\cdot5}=0.35$ ; denominator has only primes 2 and 5, so it terminates. (1)

Q3. (b). Naturals sit inside integers inside rationals inside reals. (1)

Q4. (a) $8$ . $7+3-2=8$ . (1)

Q5. (c) $12$ . $48=36\cdot1+12$ , $36=12\cdot3$ , so $\gcd=12$ . (1)

Q6. (c) $22$ . $22=4\cdot5+2$ , remainder 2. (1)

Q7. (b). Fermat: $a^{p-1}\equiv1\pmod p$ . (1)

Q8. (c) 2 and 3. $6=2\cdot3$ with $\gcd(2,3)=1$ . (1)

Q9. (b) Bézout's identity. (1)

Q10. (c) $\sqrt2$ . $\sqrt{16}=4$ , $0.\overline3=\frac13$ , $\frac{22}7$ rational. (1)

Section B

Q11. (1 mark each)

(i)–(c) $\sqrt2$ irrational
(ii)–(a) $\frac18=0.125$ terminating
(iii)–(d) $17\bmod4=1$
(iv)–(e) $\frac23=0.\overline6$ repeating
(v)–(b) $99$ : digit sum 18, divisible by 3 and 9

Section C (1 verdict + 1 justification)

Q12. TRUE. Digit sum $1+2+3+4+5+6=21$ , divisible by 3, so $123456$ is. (1+1)

Q13. TRUE. Any integer $n=\frac n1$ , a ratio of integers with nonzero denominator, hence rational. (1+1)

Q14. FALSE. $\gcd=1$ means coprime, not prime; e.g. $\gcd(8,9)=1$ but neither is prime. (1+1)

Q15. TRUE. $7-(-3)=10\equiv0\pmod{10}$ . (1+1)

Q16. FALSE. $\frac{22}7$ is only an approximation; $\pi$ is irrational (non-terminating, non-repeating). (1+1)

Q17. FALSE. $73=7\cdot10+3$ , remainder 3; not divisible by 7. (1+1)

Q18. TRUE. For coprime moduli the CRT gives a unique solution mod $mn$ . (1+1)

[
  {"claim":"90728 divisible by 11 (alt sum 22)","code":"result = (90728 % 11 == 0)"},
  {"claim":"gcd(48,36)=12","code":"result = (gcd(48,36) == 12)"},
  {"claim":"7/20 terminates: equals 0.35","code":"result = (Rational(7,20) == Rational(35,100))"},
  {"claim":"7 congruent -3 mod 10","code":"result = ((7 - (-3)) % 10 == 0)"},
  {"claim":"73 not divisible by 7","code":"result = (73 % 7 != 0)"}
]