Problem of the Week
Problem A and Solution
All Boxed In
Problem
A) Using the values below, arrange them in the six boxes to make the smallest sum and the
largest sum.
B) Using the numbers below, arrange them in the six boxes so the sum that is given is correct.
How many different solutions can you find?
Solution
A) The strategy for picking the smallest or largest number comes from realizing
that the hundreds digit has the most influence on the size of a number, and
the ones digit has the least influence on the size of the number.
When we are choosing the digits for the smallest sum, we want the smallest
numbers in the hundreds column the next smallest numbers in the tens
column and the largest numbers in the ones column. So, in our answer we
want 3 and 4 in the hundreds column, 5 and 6 in the tens column, and 7 and
9 in the ones column.
One possible solution is:
Now, since with addition, the order of the operands does not affect the sum,
we can arrange these digits in several different ways. In each column, there
are two ways of placing the digits. Since there are two ways to arrange the
digits in the hundreds column, and for each of those arrangements there are
two ways to arrange the digits in the tens column, and for each of those
arrangements, there are two ways to arrange the digits in the ones column,
there are a total of 2 ⇥ 2 ⇥ 2 = 8 solutions. They are:
357 + 469, 457 + 369, 367 + 459, 467 + 359, 359 + 467, 459 + 367, 369 + 457,
and 469 + 357.
In all arrangements, the sum is 826.
We can do similar work to find the 8 possible combinations that form the
largest sum:
964 + 753, 764 + 953, 954 + 763, 754 + 963, 963 + 754, 763 + 954, 953 + 764,
and 753 + 964.
Again, the sum in all cases is the same: 1717.
B) One possible solution is:
We need some logic to narrow down the choices in this example. Since we
know the ones digit of the sum is 8, then the ones digits for each of the two
addends must have a total of 8 or 18. There are three possible combinations
of numbers that would give us that sum: 1 + 7, 2 + 6, and 3 + 5. Since none
of these sums gives us 18, we do not have to worry about carrying a 1 into the
tens column. Looking at the tens column, we need two numbers that add up
to 6 or 16. There is only one way to do that with the numbers we are given:
1 + 5. This means that we cannot use the combinations involving 1 and 5 for
the ones column. That leaves us with the 2 + 6 sum in the ones column. The
remaining digits are 3 and 7. We have to hope that they will work for the
hundreds column. Since there is no carry from the tens column, the sum of
the digits in the hundreds column must be 10, and 3 + 7 = 10. This works!
Similar to part A, since for each column of the sum there are two possible
arrangements of the digits, there are 8 possible solutions:
312 + 756, 712 + 356, 352 + 716, 752 + 316, 316 + 752, 716 + 352, 356 + 712,
and 756 + 312.
Teacher’s Notes
As noted in the solution, the order of the operands when we are adding, does not
affect the sum. This property of addition is called the Commutative Property.
Addition and multiplication are commutative operators, but subtraction and
division are not commutative.