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Find the smallest four digit number that leaves reminder 7 when divided with 12,15,18 |
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Answer» Answer: n=16a+8 n=18b+8 Lets rearrange the EQUATION for further convenience. You’ll figure out why it is needed. n−8=16a n−8=18b From the REARRANGED equations it can be inferred that n−8 is divisible by both 16 and 18 . This doesn’t happen unless n−8 is a MULTIPLE of LCM of 16 and 18 . LCM of 16 and 18 , GIVES us second such number (read n− 8), which is divided by both 16 and 18 . [First such number is 0 itself.] We know that LCM(16,18)=144 , which is nothing but n−8 as per our discussion, giving us n as 152 . We can recheck the conditions given in the question. Reminder when 152 is divided by 16 and 18 is indeed 8 . Since this is a 3−digit number, 152 is the not the number we are looking for. Explanation: |
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