Problem 19. The cards shown 2, 3, 4, 5, 6 are placed into 2 boxes. The sums of the numbers in each box are the same. Which number must be in the box of the number 4? A, only 3, B, only 5, C, only 6, D, only 5 or 6, or E, it is impossible to know. We are dividing the numbers into 2 boxes and each of the boxes has the same sum. So all the cards are in these 2 boxes and what are in the other box? The sum of the numbers in the 2 boxes is the same as the sum of these numbers. If we add these numbers together, 2 plus 3 is 5 plus 4 makes 9 plus 5 is 14 and 14 plus 6 is 20. So 2 boxes together will have 20 in it and since the sum in each box is the same, the sum in each box is half of 20 or 10. The problem asks which number will be in the box with the number 4? So we need to figure out how we make a sum of 10 including the number 4. An obvious way is 4 plus 6 which makes 10. Now if we try adding 4 and 5, we get 9 but we don't have a 1 so we cannot use those 2 numbers and another number to make 10. Also if we add 4 and 3, we get 7, we need another 3 and we don't have that so we can't use the 3. And 4 and 2 would make 6 so we need another 4 but we only have one 4. So there is no way to make a 10 out of these numbers if we need to include the number 4 other than to add the number 6 to it. So the number 6 has to be in the box with the number 4 and then the numbers 2, 3 and 5 will be in the other box. So the answer is C. Only the number 6 will be in the box with the number 4.

number partitioning

equal sum boxes

combinatorics

problem-solving

math puzzle