The smallest number which when added to 10000 becomes divisible by 20, 24 and 30 is

Q: What is the answer to CAPF 2021 Paper I Q78?

C. Correct — C, 80. A number divisible by 20, 24 and 30 must be a multiple of their least common multiple. Factorising, 20 = 2² × 5, 24 = 2³ × 3 and 30 = 2 × 3 × 5, so the highest power of each prime gives 2³ × 3 × 5 = 120. The question is therefore how far 10,000 sits below the next multiple of 120. Dividing, 120 × 83 = 9,960, which is 40 short of 10,000, so the next multiple up is 120 × 84 = 10,080. Adding 80 to 10,000 reaches it, and no smaller addition can, since the multiples of 120 are 120 apart. (Provisional — pending the official UPSC key.)

(a)120
(b)800
(c)80
(d)83

Answer

Why

Correct — C, 80. A number divisible by 20, 24 and 30 must be a multiple of their least common multiple. Factorising, 20 = 2² × 5, 24 = 2³ × 3 and 30 = 2 × 3 × 5, so the highest power of each prime gives 2³ × 3 × 5 = 120. The question is therefore how far 10,000 sits below the next multiple of 120. Dividing, 120 × 83 = 9,960, which is 40 short of 10,000, so the next multiple up is 120 × 84 = 10,080. Adding 80 to 10,000 reaches it, and no smaller addition can, since the multiples of 120 are 120 apart.

Why the others are wrong

(a)120 — The least common multiple itself, offered because it is the number the working produces on the way. Adding 120 to 10,000 gives 10,120, which is not a multiple of 120 — and even if it were, it would not be the smallest such addition.
(b)800 — 10,800 is indeed divisible by 120, so this addition works, but the question asks for the smallest and 80 is ten times smaller.
(d)83 — 83 is the quotient when 10,000 is divided by 120, not a remainder or a shortfall. It is the number you write down if you stop the division one step early.

Concept

The least common multiple of a set of numbers is the smallest number divisible by all of them, and it is built by taking the highest power of every prime that appears in any of them. Every common multiple is a multiple of the LCM, which is what makes the LCM the right tool whenever a question asks for a number divisible by several others at once.

Once 120 is in hand the rest is a remainder calculation. Dividing 10,000 by 120 leaves a remainder of 40, and the amount that must be added is the modulus minus that remainder, 120 − 40 = 80. The mirror question — the smallest number that must be subtracted — is answered by the remainder itself, 40, giving 9,960. Reading which of the two is wanted is often the only real difficulty in this family of questions.

Key facts

LCM(20, 24, 30) = 2³ × 3 × 5 = 120.
10,000 divided by 120 leaves a quotient of 83 and a remainder of 40.
The amount to add is the divisor minus the remainder: 120 − 40 = 80.
The amount that would have to be subtracted instead is the remainder itself, 40.
The nearest multiples of 120 on either side of 10,000 are 9,960 and 10,080.

One least common multiple, one remainder

Factorise→20 = 2² × 5, 24 = 2³ × 3, 30 = 2 × 3 × 5

Least common multiple→2³ × 3 × 5 = 120

Divide 10,000 by 120→quotient 83, remainder 40

Add to reach the next multiple→120 − 40 = 80, giving 10,080

Subtract instead→40, giving 9,960

The quotient 83 and the modulus 120 are both offered as options, which is what makes stopping early expensive.

Study next

Least common multiple and highest common factor by prime factorisationThe relation HCF × LCM = product of two numbersRemainder problems and the smallest number leaving a given remainderDivisibility rules for 3, 4, 8, 9 and 11Problems on bells ringing together and lights flashing together, which are LCM in disguise

Common traps

Offering the least common multiple itself as the amount to be added.
Confusing the quotient with the remainder.
Answering the subtraction version of the question when the addition version was asked.

A least-common-multiple item dressed as a word problem. Find the LCM, divide, and read the stem again before choosing.

Related PYQs

UPSC_2010_GS1_Q712010Open & attempt →

How many numbers from 0 to 999 are NOT divisible by either 5 or 7?

(a) 313
(b) 341
(c) 686
(d) 786

Answer(c) 686

Divisibility used for counting instead of for construction. Its solution needs the multiples of 5, of 7 and of their product 35, so the same idea of a common multiple governs both questions.

Practice

practice — not a real PYQ
The smallest number that must be subtracted from 10,000 to make it divisible by 20, 24 and 30 is
- (a)20
- (b)40
- (c)80
- (d)120
Answer(b) 40 — the remainder when 10,000 is divided by their least common multiple, 120.
practice — not a real PYQ
Three bells ring at intervals of 6, 8 and 12 minutes. If they ring together at 9 a.m., they will next ring together at
- (a)9:12 a.m.
- (b)9:24 a.m.
- (c)9:36 a.m.
- (d)10:00 a.m.
Answer(b) 9:24 a.m. — the least common multiple of 6, 8 and 12 is 24 minutes.