SOLUTION: 'I have a number of cupcakes. I can pack them in boxes which contain four cakes, three cakes or eight cakes. In each case I will fill all of the boxes with none left over. What is

Algebra ->  Customizable Word Problem Solvers  -> Misc -> SOLUTION: 'I have a number of cupcakes. I can pack them in boxes which contain four cakes, three cakes or eight cakes. In each case I will fill all of the boxes with none left over. What is      Log On

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Question 1022166: 'I have a number of cupcakes. I can pack them in boxes which contain four cakes, three cakes or eight cakes.
In each case I will fill all of the boxes with none left over. What is the least number of cupcakes I could have?'

Found 2 solutions by Theo, Edwin McCravy:
Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
it looks like you are looking for the least common multiple.
start with the biggest box you have.
that's a box that contains 8 cupcakes.
so start with 8.
look to see if 3 and 4 both divide equally into it.
4 does, but 3 doesn't.
go to the next multiple of 8, which is 16.
4 divides into it evenly but 3 doesn't.
go to the next multiple of 8 which is 24.
both 4 and 3 divide equally into.
looks like the lest number of cupcakes you can have are 24.

Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!
'I have a number of cupcakes. I can pack them in boxes
which contain four cakes, three cakes or eight cakes.
In each case I will fill all of the boxes with none
left over. What is the least number of cupcakes I
could have?'
The other tutor is right that you need the least common
multiple of 4,3,and 8.  It's the same as finding the 
least common denominator when adding fractions.

Here is an easy way to find the least common multiple,
or the least common denominator when adding fractions:

List them in a row

     4  3  8

Think of any PRIME number that will divide evenly into
one or more of them.  I thought of the PRIME number 2, 
for 2 will divide evenly into both the 4 and the 8.
So we put a vertical line and a 2 on the left.  We 
divide the 4 and the 8 by the 2, and put what we get,
2 and 4 underneath the 4 and 8.  Then we just bring down 
the 3, which 2 doesn't divide evenly into, like this:

    2|4  3  8
      2  3  4

Now think of a PRIME that will divide into at least
one of those 2,3,and 4.  I thought of 3, for 3 will 
divide evenly into the 3. So we put a vertical line 
and a 3 on the left.  We divide the 3 by the 3 getting 
1, and we put the 1 under the 3 and bring down the 2 
and the 4, which 3 doesn't divide evenly into, 
like this:

    2|4  3  8
    3|2  3  4
      2  1  4

Think of any PRIME number that will divide evenly into
one or more of them.  I thought of the PRIME number 2
again, for 2 will divide evenly into both the 2 and 
the 4. So we put a vertical line and a 2 on the left. 
We divide the 2 and the 4 by the 2 getting 1 and 2 and 
we put the 1 and 2 under them and bring down the 1, 
which 2 doesn't divide evenly into, like this:

    2|4  3  8
    3|2  3  4
    2|2  1  4 
      1  1  2

Now think of a PRIME that will divide into at least
one of those 1,1,and 1.  I thought of 2, for 2 will 
divide evenly into the 2. So we put a line as a
vertical line and a 2 on the left.  We 
divide the 2 by the 2 getting 1, and we put the 1
under the 2 and bring down the two 1's, 
which 2 doesn't divide evenly into, like this:    

    2|4  3  8
    3|2  3  4
    2|2  1  4 
    2|1  1  2
      1  1  1

Now we are done for we have only 1's on the bottom
Now multiply the PRIME numbers on the left of the
vertical lines:

2∙3∙2∙2 = 24.

That method will always give the LCM (least common
multiple).  You always must divide by a prime
number, and bring down the numbers that prime will
not divide into.  You are done when there are only
1's on the bottom.  Then you multiply together all
the prime numbers on the left.

Edwin