SSC CPO practice question
Find the least number which when divided by 5, 6, 3 and 4 leaves a remainder 2, but when divided by 7 leaves no remainder.
From a SSC CPO General Awareness practice set · General Awareness
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Correct answer: (D) 182
The LCM of 5,6,3,4 is 60, so the number is of the form 60k+2. Testing divisibility by 7: 60k+2 ≡ 0 (mod 7) gives k=3 as the smallest solution, so the number = 60×3+2 = 182.
हिंदी में प्रश्न
वह न्यूनतम संख्या ज्ञात करें जिसे 5, 6, 3 और 4 से विभाजित करने पर 2 शेष बचता है, लेकिन 7 से विभाजित करने पर कोई शेष नहीं बचता।
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सही उत्तर: (D) 182
5,6,3,4 का LCM 60 है, अतः संख्या 60k+2 के रूप में होगी। 7 से विभाज्यता जांचने पर: 60k+2 ≡ 0 (mod 7) से सबसे छोटा हल k=3 मिलता है, अतः संख्या = 60×3+2 = 182 है।
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