DADS Rule!

We start with a 10 × 10 grid numbered sequentially and colour the multiples of 11. As you can see, they occur diagonally and, up to 100, the digits repeat.

As we look at multiples of 11 which are greater than 100, the pattern of repeating digits changes. And to find the more general pattern, look at any other diagonal parallel to the colored one in the 10 × 10 grid. If we take the diagonal starting with 3 say, we notice the numbers 3, 14, 25, 36, 47, 58, 69, 80. It takes only a moment to notice that there is a pattern in the difference between the units digit and the tens digit. In this diagonal, except for the last difference (which is −8), there is a constant difference of 3. If we take the diagonal beginning with 2, we get the differences as 2 and −9 respectively. We can observe that the two integers we get as the digit differences for any diagonal are 11 apart from each other. Now notice the difference between the units and tens digits for the diagonal of multiples of 11; we see that this difference is zero for all multiples of 11 less than 100.

This is a good time to take a look at the 3-digit numbers. We get the multiples of 11 as 110, 121, 132, 143 … Interestingly, the sum of the units digit and the hundreds digit less the tens digit is 0 till we hit 209. There the difference is 11. It seems like it is time to zoom in on the multiples of 11 only. To do this, we advise the use of any spreadsheet (we used Excel) to generate these rows of multiples of 11. If necessary a teacher can always take a printout of these in class. There the students can look into the patterns and color numbers with (U + H) − T = 11.

Look at Fig. 2. Here we find an interesting triangle of 209, 308, 407 … 902 which yields a difference of alternate digit sums of 11 while every other multiple of 11 gives a zero as the difference of the alternate digit sums.

What about 4 digit multiples of 11? Look at Fig. 3. We observe (U + H) − (T + Th) = 0 or 11 or −11. At this point, we decided to name this difference DADS (Difference of Alternate Digit Sums). It is interesting to notice how the triangles with DADS 11 start off as fairly large triangles but then shrink to just one number 7909. Similarly the numbers with DADS −11, appear as triangles with the same orientation initially, but then change orientation and seem to wrap across the ends of the grid.

What if we change the number of columns in the hope of finding some better pattern? We tried using 9 columns, i.e., the 1^st row become 11 … 99, 2^nd row 110 … 198, etc. Immediately, we were rewarded by a clearer pattern of triangles. See Fig. 4. The DADS 11 triangles shrink and make room for the DADS −11 ones which increase till they cover almost entire rows.

At this point, we would like to move from narrative mode to posing a few questions which follow the Low Floor High Ceiling Pattern.

Questions for Investigation

1. Are there numbers with DADS equal to 22?
2. Which is the smallest number with DADS equal to 22?
3. Are there numbers with DADS equal to −22? Which is the smallest number with DADS equal to −22?
4. What will be the smallest number with DADS equal to \(11n\)? How many digits will this number have?
5. Find a general formula for the number of digits of the smallest number with DADS equal to \(11n\) for a given value of \(n\).
6. For all multiples of 11, will the DADS be a multiple of 11?

Teacher’s Notes:

1. Students may construct numbers of the form 1010101010101010101010101010101010101010101 to obtain a number with DADS 22. From there it will be a matter of time before they start shortening the number to 2020202020202020202020 and seeing if they can get smaller numbers.
2. This is a good chance for students to proceed systematically in an investigation. Following the reasoning in step 1, they obtain the significantly shorter number 909 which has a DADS of 18 (it is not a multiple of 11). To get a DADS of 22, the number will have to be 40909.
3. This is a very interesting variation- following the reasoning in the steps above, we see that 409090 is a number with a DADS of −22. Is this the smallest number? Clearly, if there are non-zero digits in the places alternating with the units place, then the digits in the other places will have to be larger so that the difference remains as −22. Students may notice that the smallest number with a negative DADS will have an even number of digits and the smallest number with a positive DADS will have an odd number of digits. A table recording the smallest number with a particular DADS value will help them make this observation.
4. Going forward, we can ask what will be the number of digits of the smallest number with DADS \(11n\) for any natural umber. The reasoning is exactly the same as before. E.g. the smallest number with 55 as DADS should have six 9s in every alternate place starting with the units digit and then a 1 in the leading digit (55 ÷ 9 = 6 with remainder 1). It will, therefore be the 13 digit number 1090909090909. So the general formula for the smallest number with a DADS of \(11n\) will be as follows: if \(q\) and \(r\) are natural numbers such that \(11n \ ÷ \ 9 = q\) with remainder \(r\) (i.e., \(r \lt 9\)), then, the number will be
   

   \( r \times 10^{2q} \ + \ 9 \times \left( 10^{2\left(q-1 \right)}\ + \ 10^{2\left (q-2 \right)}) \ + \ … \ + 1 \right)\).

   It will be a number with \(2q + 1\) digits. We call this the DADS Rule!

5. Proof that if N is a multiple of 11, then DADS is also a multiple of 11, and vice versa:
   Let us take any number N with \((2n + 1)\) digits as

   \(a_{0} + 10a_{1} + 100a_{2} + … + 10^{2n} \times a_{2n}\)

   The alternate digit sums are \(a_{0} + a_{2} + … + a_{2n}\) and \( a_{1} + a_{3} + … + a_{2n-1}\)
   And therefore DADS for N is

   \(\left( a_{0}+ a_{2} +… + a_{2n} \right) \ – \left( a_{1}+ a_{3} +… + a_{2-1} \right)\)

   Let us consider N − DADS which is

   \(a_{0} + 10a_{1} + 100a_{2} + … + 10^{2n}\times a_{2n} -\ [ \left( a_{0} + a_{2} + … + a_{2n} \right) \ – \left(a_{1} + a_{3} + … + a_{2n-1} \right)]\)\( = a_{0} + 10a_{1} + 100a_{2} + … + 10^{2n} \times a_{2n}\ – \left( a_{0} + a_{2} + … + a_{2n} \right)+\left(a_{1} + a_{3} + … + a_{2n-1} \right)\)\( = 11a_{1} + 99a_{2} + 1001a_{3} + 9999a_{4} + … + \left( 10^{2n-1} + 1 \right)a_{2n-1} + \left( 10^{2n} – 1 \right)a_{2n}\)\( = ∑^{n}_{k=1} [\left( 10^{2k-1} + 1 \right)a_{2k-1} + \left( 10^{2k} – 1 \right)a_{2k}]\)

Now \(10^{2k – 1} + 1 = \left( 10 + 1 \right) \left( 10^{2k-2} \ – \ 10^{2k – 3} + … + 1 \right) = 11b\)for some natural number \(b\), i.e., \(10^{2k – 1} + 1\) is divisible by 11. [This step of successive decomposition may be easier for students to understand if we use an example say

which is a multiple of 11.

From this step, students may find it easier to generalise. They could also investigate if \(10^{n} + 1\) is a multiple of 11 for all \(n\) or only odd \(n\).]

Similarly \(10^{2k} \ – \ 1 = \left( 10^{2} \right)^{k} \ – \ 1 = 100^{k} \ -\ 1 = \left(100 – 1 \right) \left( 100 ^{k – 1} + … + 1\right) = 99c\) for some natural number \(c, 10^{2k} \ – \ 1\) is divisible by 99 and hence by 11.

Since N − DADS is divisible by 11, either both N and DADS are divisible by 11 or neither one is; so if DADS is a multiple of 11 so is the original N, and if DADS is not, neither is N.

Conclusion

Mathematical investigations are perfect for Low Floor High Ceiling activities. Here, we have described how a simple pattern can be recognized, investigated, played with and generalized. If your students have enjoyed DADS Rule, do let them try the same strategies with other number patterns; we hope they rule!

And don’t forget to share your students’ findings with At Right Angles.