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.

$35 = 5 \times 7$

$42 = 2 \times 3 \times 7$

$49 = 7 \times 7.$

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,
Penny