Find the smallest number which, when divided separately by 15, 20, 36, and 48, leaves a remainder of 3 in each case.
Find the smallest number which, when divided separately by 15, 20, 36, and 48, leaves a remainder of 3 in each case.
वह छोटी से छोटी संख्या ज्ञात कीजिए जिसे 15, 20, 36 और 48 से अलग-अलग विभाजित करने पर प्रत्येक स्थिति में 3 शेष बचे।
Detailed Solution & Logic
Correct Answer:
723
Step 1: Form equation
Let the number be $N$
Then:
$N - 3$ is divisible by 15, 20, 36, 48
So,
$N - 3 = \text{LCM of } (15, 20, 36, 48)$
Step 2: Find LCM
- $15 = 3 \times 5$
- $20 = 2^2 \times 5$
- $36 = 2^2 \times 3^2$
- $48 = 2^4 \times 3$
LCM = $2^4 \times 3^2 \times 5 = 16 \times 9 \times 5 = 720$
Step 3: Find number
$N = 720 + 3 = 723$
Final Answer: 723
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