Finding the smallest number satisfying multiple remainder conditions using LCM
Question
What is the smallest number greater than 1000 that when divided by any one of the numbers 6, 9, 12, 15, 18 leaves a remainder of 3?
Options
1063
1073
1083
1183
Explanation
To find a universal dividend leaving a constant remainder, calculate the Least Common Multiple (LCM) of the given divisors first: 6, 9, 12, 15, 18. Notice that 18 is a multiple of both 6 and 9, so we only need to find the LCM of 12, 15, 18.
LCM = 2² × 3² × 5 = 4 × 9 × 5 = 180.
Any number fulfilling the condition takes the form: N = 180k + 3. We must find the smallest N strictly greater than 1000. Test integer values for k:
1083 is the smallest target value exceeding 1000.
Answer: (c).
Question details
Year
2022
Paper
CSAT
Question
Q65
Section
Numerical Ability
Sub-topic
LCM & HCF
Type
Factual single
Difficulty
Easy
Source hint
Smallest number > 1000 divided by 6, 9, 12, 15, 18 leaves 3
Same sub-topic — other years
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