Number systems and HCF/LCM
Most questions here become short once you write each number as a product of prime powers. From that one step you can read off factors, HCF and LCM. Remainder and unit digit questions depend on patterns that repeat, so the work is to find the length of the pattern.
Formulas and when they apply
- Product of two numbers = HCF × LCMApplies when: exactly two numbers
- Number of factors of p^a × q^b × r^c = (a + 1)(b + 1)(c + 1)Applies when: p, q, r are distinct primes
- HCF uses the lowest power of each common prime; LCM uses the highest power of every prime that appears
- Divisible by 3 (or 9) when the digit sum is divisible by 3 (or 9)
- Divisible by 11 when (sum of digits in odd places) − (sum of digits in even places) is 0 or a multiple of 11
- Smallest number leaving remainder r with divisors a, b, c = LCM(a, b, c) + rApplies when: the same remainder r for every divisor, and r is smaller than each divisor
- Greatest divisor of x and y leaving remainders r and s = HCF(x − r, y − s)Applies when: the divisor must be larger than both remainders
- Unit digits of powers repeat in a cycle of length 1, 2 or 4Applies when: divide the exponent by the cycle length; a remainder of 0 means the last digit of the cycle
Worked examples
Example 1. Find the HCF and the LCM of 84 and 120.
Solution. 84 = 2² × 3 × 7 and 120 = 2³ × 3 × 5. The HCF uses the lowest powers of the common primes: 2² × 3 = 12. The LCM uses the highest powers of all primes: 2³ × 3 × 5 × 7 = 840. Check: 12 × 840 = 10,080 = 84 × 120.
Example 2. What is the unit digit of 7¹⁰³?
Solution. The unit digits of the powers of 7 are 7, 9, 3, 1 and then repeat, a cycle of length 4. 103 = 4 × 25 + 3, so the remainder is 3 and the unit digit is the third one in the cycle, which is 3.
Example 3. How many positive factors does 600 have?
Solution. 600 = 2³ × 3 × 5². Add 1 to each exponent and multiply: (3 + 1)(1 + 1)(2 + 1) = 4 × 2 × 3 = 24 factors.
Shortcuts and their limits
- To test divisibility by 6, 12 or 15, test two coprime parts instead (2 and 3, 4 and 3, 3 and 5).Does not apply when: Use only pairs with no common factor. A number divisible by both 2 and 4 is not necessarily divisible by 8.
- For the unit digit of a power, only the last digit of the base matters, and the exponent is reduced by the cycle length.Does not apply when: It gives the unit digit only, not the last two digits. Bases ending in 0, 1, 5 or 6 never change their unit digit.
- To find an HCF quickly, use repeated division: HCF(a, b) = HCF(b, remainder of a ÷ b).Does not apply when: It takes two numbers at a time; for three numbers find the HCF of two, then of that result and the third.
Common mistakes
- Using product = HCF × LCM for three or more numbers. It holds only for two numbers.
- Forgetting that a divisor leaving remainders must be larger than those remainders.
- Treating a remainder of 0 in a unit digit cycle as the digit 0 instead of the last digit of the cycle.
- Counting factors with the exponents themselves instead of exponent + 1.
- Mixing up the lowest power (for HCF) and the highest power (for LCM).
Independent practice made by EduVistaHQ with original questions. It is not an official examination paper, it is not affiliated with or endorsed by any examination body or company, and a practice score does not predict a real result.