Q168
ArithmeticNumber System, LCM/HCF & Divisibility
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
- b.301
- c.309
- d.400
Answer: (B) 301
The number of boxes must leave a remainder of 1 when divided by the LCM of 3, 4, 5 and 6 (which is 60), so it has the form $60k+1$, and must also be exactly divisible by 7. Testing $60k+1$ for divisibility by 7 gives the smallest valid value at $k=5$: $60\times5+1=301$, and $301 \div 7 = 43$ exactly.