Four bags contain a large number of 1s, 3s, 5s and 7s. Pick any 10 numbers from the bags above so that their total is 37.

Question

It is really annoying me and I want to solve it.  

Answer 1

Well, if you are really annoyed enough, then this is a “teachable moment” (teachers/tutors live for such a time).

Remember learning about positional number systems.  For example, decimal numbers have value depending on their place (right to left) in the number.   That is, there is 1’s place and 10’s place and 100’s place.  That way you can tell, for say for 325, that there are 5*1 + 2*10 + 3*100 total.

Now, instead of positions, you have bags and the total is computed by multiplying (up to a “large number”) by 1, 3, 5, or 7  and the total must be 37.  So how many different combinations of 10 or fewer digits is this?

Systematically, we could start by using as many 7’s as possible with a total of 37 or less, then add 5’s, then 3’s then 1’s until we get 37 (or we could start with 1’s, then 3’s, 5’s an 7’s) using 10 digits (and using a 0 is critically important in a positional number system).  When we use all possible of a specific digit, then we increment the next digit (much like an odometer, a gas pump, or other counter increments).

However, as mathematicians, we can let A,B,C,D be the how many of that number we picked from each of four bags and write:
        A*7 + B*5 + C*3 + D*5 = 37
        A + B + C + D = 10

And observe that:
     If A = 0, then B + C + D must = 10
     If A=1, then B + C + D must = 9
     If A = 2, then B + C + D must = 8
        …  and so on …

Also, When A=0, if B = 0, then C + D must = 10   … also, and so on.
Finally, When A = 0 and B = 0 and C = 0, then there is no D that is 10 or less which makes 37.

So, the “exhaustive iteration” (count like an odometer and try the total) method looks like this:
         Start answer counter at 0
         For A goes from 0 to 5 then                      (it cannot be larger than 5)
              For B goes from 0 to 7 then                    (7*5’s = 35)
                   For C goes from 0 to 10                   (12*3 is 36, but we only have 10 choices)
                       For D goes from 0 to 10                (again, only 10 choices)
                          If (A+B+C+D = 10) and ( (A*7 + B*5 + C*3 + D*1) = 37 ) then
                                    This is an answer; increment answer counter
                        NEXT
                   NEXT
                NEXT
            NEXT

How do like a computer program ??

It took 4 minutes to write and the time to run it was so fast that I couldn’t measure it.  The answer counter was 0!

What does that mean ???

Yikes!  It means that 10 numbers (using 1,3,5,and 7 as many times as you want) will not add up to 37.

Answer 2

Are there some additional instructions to this problem? Picking 10 of these numbers that add to 37 is impossible.

 

To show this, let

 

a = the number of 7s chosen

b = the number of 5s chosen

c = the number of 3s chosen

d = the number of 1s chosen.

 

Note that a,b,c and d are all integers, since you presumably can't take half of a number. We need the ten numbers chosen to add to 37, which means they satisfy the equation

 

7a+5b+3c+d=37 (7 times the number of 7s chosen plus 5 times the number of 5s chosen plus 3 times the number of 3s chosen plus the number of ones chosen)

 

They also have to satisfy

 

a+b+c+d=10, since we are choosing 10 numbers. We can manipulate this equation to get

 

d=10-a-b-c

 

If we substitute this back into the first equation, we have

 

7a+5b+3c+10-a-b-c=37

 

Simplifying, we get

 

6a+4b+2c=27

 

Now, we can factor a 2 out on the left hand side:

 

2(3a+2b+c)=27

 

Look at what we have here. Since a,b and c are all integers, 3a+2b+c is also an integer. That means that 2(3a+2b+c) is an even number, and cannot be equal to 27, since 27 is an odd number. 

 

The conclusion is that this system has no integer solutions, so it is impossible to pick 10 numbers from these bags that add to 37.

 

I think that either there is a misprint in your book or there are instructions to either solve the problem or determine that it is unsolvable. Check whether there are are instructions like that and ask your teacher for clarification if you can't find them.

Answer 3

It's annoying because there is no solution.

 

They want you to pick 10 numbers, which is 5 sets of 2 numbers, that add to 37.  But all the numbers are odd, and when you add together two odd numbers, you get an even number.  So every one of the 5 sets of 2 numbers is going to add to an even number.  When you add any number of even numbers together, you'll still end up with an even number, so the 5 sets of 2 odd numbers are going to add to an even number. Since 37 is an odd number, there's no way that the 10 numbers can add up to 37.