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RailwaysSSCMathHCF AND LCMMedium

HCF AND LCM practice question for Railways, SSC. Review the correct answer and explanation, then continue with related practice below.

Question language / प्रश्न भाषा

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 Answerसही उत्तर

C. 67C. 67

Explanationव्याख्या

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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