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Problem of the Week
Problem B and Solution
That’s About Right

Problem

  1. Place the digits \(1\), \(3\), \(6\), \(7\), \(8\), and \(9\) in the boxes shown so that each box contains a different digit, and the sum is as close as possible to \(99\).

    Six boxes are arranged to represent the digits in an addition problem written vertically. A two-digit integer is added to another two-digit integer and the result is also two-digit integer.

  2. The digits \(5\), \(6\), and \(8\) have been placed in three of the boxes shown. Place the digits \(0\), \(1\), \(2\), \(3\), \(4\), \(7,\) and \(9\) in the remaining boxes so that each box contains a different digit, and the sum is as close as possible to \(1000\).

    Ten boxes are arranged to represent the digits in an addition problem written vertically. A three-digit integer with tens digit 8 is added to a three-digit integer with hundreds digit 5. The result is a four-digit integer with ones digit 6.

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Solution

  1. Notice that \(98\) is the closest number to \(99\) that could possibly be formed using the digits \(1\), \(3\), \(6\), \(7\), \(8,\) and \(9\). Let’s see if we can arrange the remaining digits, \(1\), \(3\), \(6,\) and \(7\), to get a sum of \(98\).

    Using the digits \(1\), \(3\), \(6,\) and \(7\), to get a sum with a ones digit of \(8\), we must place the \(1\) and the \(7\) in the two boxes in the ones column. If we place the remaining digits, \(3\) and \(6\), in the tens column, we will get a sum with a tens digit of \(9\). Therefore, it is possible to arrange the digits to get a sum of \(98\), which is the closest possible sum to 99. We also see that there are four possible ways to arrange the digits in the boxes to produce this sum.

    The digits are placed in the boxes in four different ways. This gives four addition problems: 61 plus 37, 67 plus 31, 31 plus 67, and 37 plus 61. The result of each addition problem is 98.   

  2. Given the digits \(0\), \(1\), \(2\), \(3\), \(4\), \(7,\) and \(9\), along with the placement of the \(6\), the closest number to \(1000\) that could be formed is \(0976\). The next closest number is \(1026\).

    We will first see if we can place the digits to get a sum of \(0976\). If the sum is \(0976\), then the digits \(0\), \(9\), and \(7\) have been placed, and the remaining boxes will be filled with the digits \(1\), \(2\), \(3,\) and \(4\). Of these digits, the only two that have a sum with ones digit \(6\) are \(2\) and \(4\). Therefore, the \(2\) and the \(4\) would need to go in the ones column. This leaves \(1\) and \(3\) to be placed. Looking at the tens column of the sum, we need to place one of these numbers in the tens column so that the sum of that number with 8 has a ones digit of \(7\). This is not possible. Therefore, we see that it is not possible to place the numbers so that the sum is \(0976\).

    Next, we try to place the digits to get a sum of \(1026\). If the sum is \(1026\), then the digits \(0\), \(1\), and \(2\) have been placed, and the remaining boxes must be filled with the digits \(3\), \(4\), \(7,\) and \(9\). Of these digits, the only two that have a sum with ones digit \(6\) are \(7\) and \(9\). Therefore, the \(7\) and the \(9\) would need to go in the ones column. This leaves \(3\) and \(4\) to be placed. Looking at the tens column of the sum, we need to place one of these numbers in the tens column so that the sum of that number with \(8\) and \(1\) (the carry from the ones column) has a ones digit of \(2\). This is possible if we place the \(3\) in this box. That leaves the \(4\) to go in the empty box in the hundreds column. Indeed, we see that the sum of \(4\) with \(5\) and \(1\) (the carry from the tens column) is \(10\), as required.

    Thus, it is possible to arrange the digits to get a sum of \(1026\), and this is the closest we can get to a sum of \(1000\). We see that there are two possible ways to arrange the digits in the boxes to produce this sum.

       

Extension: Can you find a better solution for (b) by placing the \(5\), \(6,\) and \(8\) elsewhere?