1. Which consecutive pair has LCM equal to 210?
Consecutive numbers are co-prime; LCM is their product. 14×15 = 210. Other pairs yield products not equal to 210.
LCM(14,15) = 14×15 = 210
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Aptitude & Reasoning · Question Set
HCF–LCM & Factors interview questions for placements and exams.
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Aptitude & Reasoning
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Consecutive numbers are co-prime; LCM is their product. 14×15 = 210. Other pairs yield products not equal to 210.
LCM(14,15) = 14×15 = 210
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Scale by 10: gcd(18,24)=6. Scale back by 10 → 0.6. Scaling makes decimal HCF problems straightforward.
gcd(18,24)=6 ⇒ 6/10 = 0.6
For positive integers a, b: a×b = HCF×LCM. Thus LCM = 630 ÷ 9 = 70. This identity is very useful in quick calculations.
LCM = (a×b)/HCF = 630/9 = 70
1000 ÷ 18 gives remainder 10 (since 18×55=990). To reach a multiple, add 8. So 1000 + 8 = 1008, which is divisible by 18.
1000 mod 18 = 10 ⇒ need 18−10 = 8
If N leaves remainder 3 with all, then N−3 is divisible by LCM(4,5,6)=60. So least N = 60 + 3 = 63.
LCM(4,5,6)=60 ⇒ N=60k+3 → k=1 ⇒ 63
If a number d leaves the same remainder, it divides differences. Here 175−130 = 45. The greatest such d is 45. Thus d = 45 satisfies the condition.
d | (175−130)=45 ⇒ max d = 45
If N leaves remainder 7 with all, then N−7 is divisible by LCM(12,18,21). LCM = 2²×3²×7 = 252. Least N = 252 + 7 = 259.
LCM = 252; N ≡ 7 ⇒ N = 252k+7 → k=1 ⇒ 259
For any positive integers, product of HCF and LCM equals product of the numbers. This follows from prime-power min–max properties.
h×l = a×b
Use Euclid’s algorithm: 210 mod 84 = 42, and 84 mod 42 = 0. Hence HCF is 42. HCF is the greatest number dividing both values without remainder.
gcd(210,84) → gcd(84,42) → gcd(42,0) = 42
Compute stepwise: gcd(96,144) = 48; then gcd(48,240) = 48. So the greatest common divisor of all three is 48.
gcd(96,144)=48 → gcd(48,240)=48
LCM of fractions = LCM of numerators ÷ HCF of denominators. LCM(3,7)=21; HCF(20,15)=5. So LCM = 21/5.
LCM(3/20,7/15) = LCM(3,7)/gcd(20,15) = 21/5
Prime factors: 18 = 2×3², 24 = 2³×3, 30 = 2×3×5. Take highest powers: 2³×3²×5 = 360. LCM uses the maximum exponent of each prime across the numbers.
LCM = 2^3 × 3^2 × 5 = 360
a×b = HCF×LCM ⇒ other = (13×455)/91. Since 91=13×7, we get 455/7 = 65. So the second number is 65.
other = (13×455)/91 = 65
Count(6) = ⌊1000/6⌋=166; Count(8)=125; Count(LCM 24)=41. Use inclusion–exclusion: 166+125−41=250. Hence 250 integers.
166 + 125 − 41 = 250