वह संख्या ज्ञात कीजिए जिसे क्रमागत रूप से 5, 7 और 9 से विभाजित करने पर क्रमशः 2, 0 और 2 शेषफल बचते हैं, और अंतिम भागफल 15 है।

This question was previously asked in
MPPGCL JE Electrical 28 April 2023 Shift 3 Official Paper
View all MPPGCL Junior Engineer Papers >
  1. 4797
  2. 4977
  3. 4597
  4. 4687

Answer (Detailed Solution Below)

Option 1 : 4797
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Detailed Solution

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दिया गया है:

भाजक: 5, 7, 9

शेषफल: 2, 0, 2

अंतिम भागफल: 15

प्रयुक्त सूत्र:

भाज्य = (भाजक × भागफल) + शेषफल

गणनाएँ:

तीसरा भाग: 9 × 15 + 2 = 137

दूसरा भाग: 7 × 137 + 0 = 959

पहला भाग: 5 × 959 + 2 = 4797

इसलिए, वह संख्या 4797 है।

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