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Question

# A man has a certain number of small boxes to pack into parcels. If he packs 3,4,5 or 6 in a parcel, he is left with one, if he packs 7 in a parcel, none is left over. What is the number of boxes, he may have to pack?

A
106
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B
301
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C
309
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D
400
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Solution

## The correct option is D. 301Given, if the person packs 7 boxes then none of the box left out. So, the total number of boxes would be multiple of 7.It is given that, if he packs 3,4,5 or 6 boxes in a parcel he is left with one box.So, we need find a number that is multiple of 7 and when it divided by 3,4,5 or 6, it leaves remainder of 1.Lets find the LCM of 3,4,5 and 6.Thus, L.C.M of 3,4,5 and 6 is 3×2×2×5=60Since, the required number leaves the remainder 1 when it is divisible by 3,4,5,6 and no remainder when it is divisible by 7.Therefore the required number is of the form 60x+1⇒60x+1 is a multiple of 7Now, lets check for 60x+1 is divisible by 7 by substituting the natural numbers sequentially,60(1)+1=61=(7×8)+3 is not divisible by 7.60(2)+1=121=(7×17)+2 is not divisible by 7.60(3)+1=181=(7×25)+6 is not divisible by 7.60(4)+1=241=(7×34)+3 is not divisible by 7.60(5)+1=301=(7×43)+0 is divisible by 7.So, the least number is 301 when divided by 3,4,5,6 leaves a remainder 1, but when divided by 7, there will be no remainder.Hence, the correct option is D. 301

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