CSAT Solved Papers/ 2024/Q19

2024 CSAT — Q19

Quant Number theory 2.5 marks Medium

A can $X$ contains $399$ litres of petrol and a can $Y$ contains $532$ litres of diesel. They are to be bottled in bottles of equal size so that whole of petrol and diesel would be separately bottled. The bottle capacity in terms of litres is an integer. How many different bottle sizes are possible?

A 3
B 4 Answer
C 5
D 6

Worked rationale

A bottle of integer capacity $c$ litres must measure out all of the petrol and all of the diesel with nothing left over, so $c$ must divide both $399$ and $532$ . Therefore $c$ is a common divisor of $399$ and $532$ , i.e. a divisor of $\gcd(399,532)$ .

Factorise (fast): $399 = 3 \times 7 \times 19$ and $532 = 2^2 \times 7 \times 19$ . The shared part is $7 \times 19 = 133$ , so $\gcd = 133$ .

The possible bottle sizes are the divisors of $133 = 7 \times 19$ :

\{1,\ 7,\ 19,\ 133\} \quad\Rightarrow\quad 4 \text{ sizes.}

Answer: (b) 4.

Why the other options miss

A

a convention slip: drops the size- $1$ bottle, assuming a “bottle” must hold more than a litre. Nothing in the question excludes $c=1$ ; an integer capacity of $1$ L is valid, so this under-counts by one.
C

an arithmetic slip: a factorisation error — e.g. treating $399 = 3\cdot7\cdot19$ but $\gcd$ taken as $3\cdot7\cdot19=399$ ‘s divisor count region, or counting divisors of a wrong gcd like $399$ ( $\tau(399)=8$ , then partly halved). Comes from not actually taking the
D

solved the wrong question: counts divisors of one can only, or sums divisors of $7$ and $19$ structures incorrectly; treats “bottle sizes” as something other than common divisors.

Specialist insight

“Equal containers measuring out two (or more) whole quantities” is always HCF — and “how many sizes” means count the divisors of the HCF, not just name the HCF. The two-step structure is what separates the gold solver from the panicked one: (1) HCF gives the largest bottle; (2) every divisor of that HCF is also a valid bottle, so the count is $\tau(\text{HCF})$ . Because $133 = 7\times19$ is a product of two distinct primes, $\tau = 2\times2 = 4$ instantly — no listing needed. The recurring trap on this template is the phrase “different sizes”: examiners pay you for realising it asks for the divisor count, and they plant the off-by-one by tempting you to exclude the $1$ -litre bottle. Unless the stem says “more than 1 litre,” keep it.

The trap, in one line

"How many sizes" = count the divisors of the HCF (here $4$ ), not the HCF itself ( $133$ ) — and don't drop the $1$ -litre bottle.

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