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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?

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Solution

The correct option is **D.** 301

Given, 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=60

Since, 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 7

Now, 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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