



 
Hi Cheska, I'll show you how to solve problem 1 and you can do the rest. If Levy can pack 35 chocolate bars into boxes so that there are no extra bars then the number of boxes must be a multiple of 35. If Levy can pack 42 chocolate bars into boxes so that there are no extra bars then the number of boxes must be a multiple of 42. If Levy can pack 49 chocolate bars into boxes so that there are no extra bars then the number of boxes must be a multiple of 49. Hence the number of boxes must be a multiple of 35, 42 and 49. One number that is a multiple of 35, 42 and 49 is $35 \times 42 \times 49 = 72030,$ but is there a smaller number. One way to approach this is to write 35, 42 and 49 as products of primes.
Now look for the smallest multiple of 35 and 42. This number is a multiple of 35 so its prime factorization must contain a 5 and a 7. Since if is also a multiple of 42 its prime factorization must contain a 2, 3 and a 7. Hence its prime factorization must contain at least one 2, one 3 one 5 and one 7. Hence the smallest multiple of 35 and 42 is $2 \times 3 \times 5 \times 7.$ Is this number also a multiple of 49? No it isn't since a multiple of 49 must have two sevens in its prime factorization. Thus to be a multiple of 35, 42 and 49 the number must be at least $2 \times 3 \times 5 \times 7 \times 7.$ This is the smallest possible number of boxes. It is called the least common multiple (lcm) of 35, 42 and 49. I hope this helps, 



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