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Find the least number which when divided by 12, 15 and 20 leaves remainder 7 in each case.

वह सबसे छोटी संख्या ज्ञात कीजिए जिसे 12, 15 और 20 से भाग देने पर प्रत्येक बार 7 शेष बचे।

Answer and explanation are free to read — no sign-in required. / उत्तर और व्याख्या पढ़ने के लिए साइन-इन आवश्यक नहीं है।

A

5757

B

6060

C

6767

✓

D

127127

Correct Answer / सही उत्तर

C. 67
67

Explanation / व्याख्या

Since the remainder is the same (7) in each case, subtract 7 from the required number.

The resulting number must be exactly divisible by 12, 15, and 20.

So, find:

LCM of 12, 15, and 20

Prime factorization:

12 = 2² × 3
15 = 3 × 5
20 = 2² × 5

Therefore,

LCM = 2² × 3 × 5 = 60

Now add the common remainder:

Required number = 60 + 7 = 67

चूँकि प्रत्येक स्थिति में समान शेषफल 7 है, इसलिए पहले 12, 15 और 20 का LCM ज्ञात करेंगे।

12 = 2² × 3
15 = 3 × 5
20 = 2² × 5

अतः,

LCM = 2² × 3 × 5 = 60

अब इसमें शेषफल 7 जोड़ेंगे:

अभीष्ट संख्या = 60 + 7 = 67

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