# Hardest math question in the world?

she sold for exactly \$100. She sold

caviar sandwiches for \$5.00 each, the

bologna sandwiches for \$2.00, and

the liverwurst sandwiches for 10 cents.

How many of each type of sandwich

did she make?

Tricky.

C = # of caviar

B = # bologna

L = # of liverwurst

C + B + L = 100

5C + 2B + L/10 = 100

multiply the 2nd equation by 10 and subtract the two

49C + 19B = 900

Now, there are lots of ways to do that, but you need both C and B to be integers. Notice that 900 is divisible by 10. Divide the whole equation by 10

\$4.9 C + \$1.9 B = \$90

Now, we know she need to make exactly \$90 selling just caviar and bologna. Each time she sells a sandwich, she'll have to give a dime in change. To get back to an exact dollar figure, she'll need to sell some multiple of 10 sandwiches. So, make the substitution

C + B = 10 n

Now, solve for C and substitute

49 n - 3 B = 90

Remembering that n and B are both integers, n must be a multiple of 3 (divide that equation by 3 and see). So try, n=3, 6, 9, etc.

For n=3, B = 19

For n=6, B = 68

For n=9, B = 117 (which is impossible.)

If B = 19, then C = 11 and L = 70.

If B = 68, then C = -8 which is impossible.

So, the only possible answer is 11 caviar sandwiches, 19 bologna sandwiches and 70 Liverwurst sandwiches.

#1

...What are you, a 1st grader?

#2

This is impossible to answer because there can be an infinite number of solutions to this problem.

18 Caviar Sandiwiches = \$90

4 Bologna Sandwiches = \$8

20 Liverwurst Sandwiches = \$2

This all equals \$100. There can be many more solutions to this problem but this is one of them.

#3

You example only equals to 42 sandwiches... might want to check the question again.

#4