Arithmetic & Number Systems

Question 1

(a) Pseudocode (2 marks):

function gcd(a, b):
    while b != 0:
        r = a mod b
        a = b
        b = r
    return a

Trace for $\gcd(1134, 420)$ (2 marks — one per correct remainder chain):

• $1134 \bmod 420 = 294$ (since $1134 = 2\cdot420 + 294$)
• $420 \bmod 294 = 126$ (since $420 = 1\cdot294 + 126$)
• $294 \bmod 126 = 42$ (since $294 = 2\cdot126 + 42$)
• $126 \bmod 42 = 0$
• $\gcd = \boxed{42}$

(b) (3 marks) $\text{LCM} = \dfrac{a\cdot b}{\text{HCF}} = \dfrac{1134\times420}{42} = \dfrac{476280}{42} = \boxed{11340}$. Why: HCF×LCM = product; dividing product by HCF must give an integer since HCF divides both factors. (1 mark for identity, 1 for arithmetic, 1 for integer check.)

(c) (4 marks) Proof that $\gcd(a,b)=\gcd(b,r)$ where $a = qb + r$:

• Let $d$ be a common divisor of $a$ and $b$. Then $d\mid a$ and $d\mid b$, so $d\mid(a-qb)=r$. Hence $d$ divides both $b$ and $r$. (1.5)
• Conversely, let $d$ be a common divisor of $b$ and $r$. Then $d\mid qb+r = a$, so $d$ divides both $a$ and $b$. (1.5)
• Therefore the sets of common divisors of $(a,b)$ and of $(b,r)$ are identical; in particular their greatest elements are equal, so $\gcd(a,b)=\gcd(b,r)$. (1)

(d) (3 marks) For $N=p_1^{e_1}p_2^{e_2}\cdots$, number of divisors $= (e_1+1)(e_2+1)\cdots$ (2 marks for formula). Here $N=2^3\cdot3^2\cdot5^1$ gives $(3+1)(2+1)(1+1)=4\cdot3\cdot2=\boxed{24}$ divisors (1 mark).

Question 2

(a) (4 marks) Unitary: 1 km needs $12/180 = 1/15$ L. For 315 km: $315\times\tfrac{1}{15}=21$ L. Answer: 21 L. (2 marks). This is direct proportion: more distance ⇒ more fuel; ratio distance/fuel constant physically (constant consumption rate). (2 marks)

(b) (4 marks) Inverse: pipes × hours = constant = $6\times8 = 48$ pipe-hours (2). With 10 pipes: hours $= 48/10 = 4.8$ h. Answer: 4.8 hours (1). The total work to fill the tank is fixed ("48 pipe-hours"), so if more pipes share it, each contributes less time; the product is conserved (1).

(c) (4 marks) Cost = 4000. Profit 17% ⇒ Selling Price $= 4000\times1.17 = 4680$ (1). SP is after 10% discount on Marked Price: $SP = 0.90\times MP$ (1). So $MP = 4680/0.90 = 5200$. Marked Price = ₹5200 (2).

(d) (2 marks) $3.0\times10^8\ \text{m/s} = 3.0\times10^5\ \text{km/s} = 300000\ \text{km/s}$. Dividing by 1000 shifts the decimal point 3 places left, reducing each digit's place value by three orders (m → km). (1 answer + 1 explanation)

Question 3

(a) (4 marks)

• Innermost: $3^2 - 3 = 9 - 3 = 6$ (1)
• $36 \div 6 = 6$; and $|4-9| = |-5| = 5$ (1)
• Bracket: $6 - 5 = 1$ (1)
• $7 + 2\times 1 = 7 + 2 = \boxed{9}$ (1)

(b) (5 marks) Squaring method:

• Both sides non-negative, so compare squares. $(|x+y|)^2 = (x+y)^2 = x^2 + 2xy + y^2$. (1)
• $(|x|+|y|)^2 = x^2 + 2|x||y| + y^2$. (1)
• Difference: $(|x|+|y|)^2 - |x+y|^2 = 2|xy| - 2xy = 2(|xy|-xy)\ge 0$ since $|xy|\ge xy$ always. (2)
• Hence $|x+y|^2 \le (|x|+|y|)^2$, and taking non-negative square roots preserves the inequality: $|x+y|\le|x|+|y|$. Equality holds iff $xy=|xy|$, i.e. $xy\ge0$ (same sign or one is zero). (1)

(c) (3 marks) $7/8 = 0.875$ (1); as percentage $= 87.5\%$ (1). For $0.\overline{6}$: let $x=0.666\ldots$, then $10x = 6.666\ldots$, subtract: $9x = 6$, so $x = 6/9 = 2/3$. (1)

[
  {"claim":"gcd(1134,420)=42","code":"result = (gcd(1134,420)==42)"},
  {"claim":"LCM(1134,420)=11340 via product/gcd","code":"result = (1134*420//gcd(1134,420)==11340)"},
  {"claim":"divisors of 2^3*3^2*5 = 24","code":"result = ((3+1)*(2+1)*(1+1)==24)"},
  {"claim":"fuel for 315km at 180km/12L is 21L","code":"result = (Rational(12,180)*315==21)"},
  {"claim":"marked price 5200 gives 17% profit after 10% discount","code":"result = (Rational(4680,90)*100==5200 and 4000*Rational(117,100)==4680)"},
  {"claim":"nested BODMAS expression equals 9","code":"result = (7+2*(36//(3**2-3)-abs(4-9))==9)"},
  {"claim":"0.666... repeating equals 2/3","code":"result = (Rational(6,9)==Rational(2,3))"}
]