Consider all the possible arrangements of a three-digit number containing each of the digits 1, 2, and 3 once only. Your students should be able to write down the six possibilities which later on in their mathematical development they will associate with:
3! = 3 x 2 x 1
Examining these six, ask your students to find out which ones have the following combined properties. Reading from the left:
* the first digit is divisible by 1;
* the number formed by the first two digits is divisible by 2; and
* the number formed by the first three digits is divisible by 3.
The answers are clearly 123 and 321 since they are the only two that form an even number with the leading two digits.
Next, ask your students to consider a similar problem with the four digits 1, 2, 3 and 4, but this time the number formed by the four digits must be divisible by 4 as well. There are now 4! = 4 x 3 x 2 x 1 = 24 different fourdigit numbers to consider. However, using the logic of the previous paragraph students can reduce their search by noting that the even digits 2 and 4 must be in the second and fourth positions in either order, and similarly with the odd digits 1 and 3 in the remaining positions.
Now the Rule of Three tells us that 4 cannot be in the second position from the left. Also, if we denote an even digit by E and an odd digit by D then the four-digit number sought has the form DEDE which is DE00 + DE. Now any product of 100 is always divisible by 4, therefore DEDE is divisible by 4 if DE is divisible by 4. Therefore, logic or a simple exhaustive method both discover that there are no solutions to this four-digit problem.
Now, extend the problem to six-digit numbers using each of the digits 1, 2, 3, 4, 5 and 6 once only, but this time the number must be divisible by 6 as well. The reason that I have skipped five-digit numbers with the digits 1,
2, 3, 4, 5 will soon be obvious. There are now 6! = 720 different numbers to check. But mathematical logic shows that the even digits can only be placed in the second, fourth and sixth positions in six different ways, the 5 has to be placed in the fifth position, and the remaining two odd digits placed in the remaining positions. This logic means that there are now only twelve numbers to check.
Furthermore, our earlier discussion of division by 4 means that the digit 4 cannot be placed in the second or fourth place from the left. So it must reside at the end. Therefore, there are now only four numbers to check, which is a substantial reduction from 720. Your students should now be able to find the only two solutions using the Rule of Three for both the first three positions and the whole six-digit number.
Now they are ready to solve the seven-digit problem using the digits 1, 2, 3, 4, 5, 6, and 7. This time the number formed must be divisible by 7 as well. The exhaustive trial and-error method is no longer an option since there are 7! = 5040 different combinations possible. However logic tells us that the form of the answer is:
D E D E 5 E D
Therefore the number of possibilities has been reduced to 3! x 3! = 36. Note that the introduction of the digit 7 means that the digit 4 could now be in the second position in numbers starting with 147 or 741. But the digit 4 still cannot be in the fourth position, and this reduces the final search to 24.
Practice with division by 7 reveals that there are no solutions.
Eight-digit and nine-digit problems
Do not despair! There is an answer for the digits 1 to 8 inclusive, and also for the digits 1 to 9 inclusive. See if your students can find them.
This article has slowly crept up on these two more difficult problems to emphasise that one way to solve difficult problems is to think of an easier-related problem and solve it first. This may then provide insight into the more difficult problem.
The final problem is to find the six-digit number that satisfies the requirements using the digits 2, 3, 4, 5, 6 and 7 once only in the number.
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|Author:||de Mestre, Neville|
|Publication:||Australian Mathematics Teacher|
|Date:||Sep 22, 2018|
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