The sum of the two digits of a two-digit number is 11. The tens digit is one more than 4 times the units digit. What are the two digits? We will solve this problem using four boards, the first two boards representing the first equality, and the second two representing the second equality.
T + U = 11 We call the tens digit T and represent it on the board by a triangle counter. We call the units digit U and represent it on the board as a square counter. These counters we will call "variable counters", or simply "variables", because their values will be found when the problem is solved. So the value on the first board is T + U, and the value on the second board is 11. The two boards together can be read as the equation T + U = 11.
T + 4U = 1 The tens digit is one more than four times the units digit. So the two boards above represent T = 4*U + 1. We want to subtract the values on the B boards from the values on the A boards in order to get rid of T and isolate U on the left side of board A. But we can’t perform the subtraction because we don’t have enough square counters on the top right A board. So we will add the same value, 4U (4 square counters) to both A boards. (The B boards don’t change.) The A boards now look like this:
T + 5U = 4U + 11
Now we can subtract the values on the B boards from those on the A boards:
5U = 10 The red and white counters on 1 above cancel, and we have 5U = 10, shown below. Dividing each side by 5, we have the value of the units digit, the square counter, is 2:
5U = 10
U = 2 To get the value of the T, we substitute the value 2 for each U on board B.
T = 9 So the tens digit, the triangle counter, is 9. 92 is our two-digit number! Let’s check it. The sum of the digits is 11: 9 + 2 = 11. The tens digit is one more than four times the units digit. 9 = 1 + 4*2. 9 = 9. Got it! Number Board index |