HCF & LCM · Topic 4 of 6

🔁 Remainder Problems

4 exam question types, fully solved

Remainder questions look fiddly but split into just two families. The trick is reading whether the answer is built from the LCM (a number with a fixed remainder) or from the HCF (the biggest divisor giving a remainder).

Two patterns to recognise

• “Least number leaving remainder when divided by several numbers” — the number has the form . Build the LCM, then pick the smallest that fits any size limit.

To make a value exactly divisible: find its remainder on division by the LCM, then subtract it (to drop to the previous multiple) or add (to climb to the next).

• “Greatest number that divides several numbers leaving the same remainder” — the remainder cancels in the differences, so the answer is of the numbers.

• “…leaving different/known remainders” — subtract each remainder from its own number first, then take the of the results.

Golden rule: never take the HCF of the raw numbers when a non-zero remainder is involved — adjust first.

⚖️HCF & LCM MachineSame prime factors — lowest powers make the HCF, highest powers make the LCM

First

Second

Third (optional)

72=2³×3²

120=2³×3×5

HCF → only the primes shared by every number (highlighted), each at its LOWEST power. LCM → every prime that appears anywhere, each at its HIGHEST power.

HCF=2³×3=24

LCM=2³×3²×5=360

Check: HCF × LCM = 24 × 360 = 8,640
a × b = 72 × 120 = 8,640 — they match ✓

This identity works for exactly two numbers — it fails for three or more.

The least number leaving remainder 2 on division by 3, 4 and 5: , so it is .

Q1The least number leaving remainder 2 when divided by 3, 4 and 5 is:

Q2The greatest number that divides 64 and 100 leaving remainder 4 each is:

Q3The smallest number divisible by 8, 12 and 16 is:

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Real exam questions — Remainder Problems

4 question types · 15 solved examples from real SSC papers

Same remainder ⇒ HCF of the differences. Known remainder ⇒ subtract it first, then HCF. “Least number leaving remainder r” ⇒ LCM·k + r.

How to solve this type

A number that leaves the SAME remainder

r

on division by several numbers has the form

\text{LCM}\times k+r

. To make a value exactly divisible, find its remainder on division by the LCM: subtract that remainder (to go down) or add (LCM − remainder) (to go up). Build the LCM once, then just pick the right

k

Find the least 4-digit number which, when divided by 2, 3, 4, 5, 6 and 7, leaves remainder 1.

A1051B1261C2101D841

Find the least number that must be subtracted from 2000 so the result is divisible by 15, 20 and 25.

A100B50C150D200

Find the least number that should be added to 478 so the result is exactly divisible by 5, 6 and 12.

A52B62C12D2

Find the greatest 4-digit number divisible by 15, 25, 40 and 75.

A9600B9000C9500D9200