Is 411281013 divisible by 11?

That simple question is what we will try and answer in this piece. Note, we aren’t looking to divide anything; we want, only, to test divisibility. There are several ways to do this, of course. We could factorize the number to see if 11 exists as a factor. We could go so far as to perform long division by 11. But that would defeat the prupose of this exercise. What we want is the simplest way to test divisibility. Ideally, we want to create a foolproof shortcut method that we can apply quickly to numbers of any length. So, a method that helps us deal quickly with only 4-digit numbers will not do.

There are many reasons we are conducting this exercise. While it would be good to know a quick method to check divisbility by 11 this is too trivial a drop in the ocean of skills we want to build. Why we are doing this is to build some other skills and concepts; skills and concepts that will help us in the long run.

One of those skills is pattern-recognition and application. We want to take what we will learn under the head of Divisibility by 11 and apply it to various other numbers. Do you remember the divisibility test for 13? How about for 19? For 73? We hope that you will take the learnings from today’s exercise and be able to not just understand other divisibility tests but be able to come up with your own.

More importantly, however, we are doing this to show how beautiful Mathematics is; how closely the various topics and disciplines within Mathematics are connected. At school we are introduced to Numbers way before we learn about Exponents and Calculus but we never revisit Numbers to see what new insights we can gain by applying the newer techniques to older knowledge.

To work, then. Let us refresh the Fundamental Test for Divisibility by 11. From the WikiHow’s entry on the topic (http://www.wikihow.com/Check-Divisibility-of-11) we get:

1. Add the digits of odd positions from left to right of the number.Do the same thing for even positions of the number from left to right.

Subtract both sums.

If the difference is 0 or a multiple of 11, then the original number is divisible by 11.

2. Let’s see this method in action with a simple example. Take the number 1364. WikiHow’s method asks us to add up the digits in the odd places from the left. Here the digits in the odd places are 1(1st digit) and 6(3rd digit). Adding we get 7.

3. Doing this again for the digits in the even places we get 3+4 (in 2nd and 4th place, respectively) again equalling 7. We have two sums. Subtracting those we get 7 – 7 = 0. According to the Divisibility Test, since we got 0, the original number, 1364, is divisible by 11.

4. Try this with 6397 and we get 15 – 10 = 5. Hence, 6397 is not divisible by 11.

That was simple enough. And it is an elegant enough test. Two questions immediately spring into the inquisitive mind. First, can we simplify this even further? Absolutely, but we will get to that in just a bit. More importantly, we need to ask, why does this happen? As in, what do the sums of alternate digits have to do with 11? In these questions lies the basis for new discovery and insight. Let us understand this test a little.

We are asked to sum the odd place digits and the even place digits separately and then subtract the two sums. So, for a number like 1364 we are effectively doing 1 + 6 – (3 + 4). Re-writing, we get 1 – 3 + 6 – 4. For longer numbers this will take the following form:

Ten Millions – Millions + Hundred Thousands – Ten Thousands + Thousands – Hundreds + Tens – Units

and so on with alternating ‘+’ and ‘-‘ signs. Writing this a tad more mathematically, for a number of the form ABCDEFGH, we get:

A.(10⁷) – B.(10⁶) + C.(10⁵) – D.(10⁴) + E.(10³) – F.(10²) + G.(10¹) – H.(10⁰) (Expression 1)

A little more manipulation yields

A.(-1⁷).(10⁷) + B.(-1⁶).(10⁶) + C.(-1⁵).(10⁵) + D.(-1⁴).(10⁴) + E.(-1³).(10³) + F.(-1²).(10²) + G.(-1¹).(10¹) + H.(-1⁰).(10⁰) (Expression 2)

While that looks very technical all it says is that, to apply the test for divisibility by 11, each digit needs to be multiplied by -1 to the power of the place value of the digit. The rest is all addition as opposed to the alternate addition and subtraction in the previous version. Expression 2 will help us in understanding how this test works.

Let us also revisit the concept of Modulo (Remainder).

A( Modulo B) is, simply, the remainder when A is divided by B. For instance, 4( Modulo 7) is 4 and 11( Modulo 7) is also 4 because the remainder when, either 4 or 11, is divided by 7 is 4. This is because both these numbers can be written in the form 7x+4. For 4, x = 0 so 7.(0) + 4 = 4. For 11, x = 1 so 7.(1) + 4 = 11.

What is of note that every number of the form 7x + 4 can also be written as 7y + (-3). If a number is 4 more than one multiple of 7 it is 3 less than the next multiple of 7. We can then say that 11( Modulo 7) is 4 or -3. Both statements are correct. These two possible values of the Modulo operation are, obviously related. 4 – (-3) = 7 – the number we were dividing with. Just to test our comprehension, what are the possible values for 13 ( Modulo 8)?

One value is 5. The other is -3. Take a minute to revise these concepts before moving on.

On to the meat of the problem, then. Let us ask a few questions to start.

What is 10⁰( Modulo 11)?

What is 10¹( Modulo 11)?

What is 10²( Modulo 11)?

.

.

What is 10³¹(Modulo 11)?

If, for 10³( Modulo 11) you actually divided 1000 by 11 let us offer a simpler way. Note that 10⁰( Modulo 11) is 1. What is 10¹( Modulo 11)? You might be tempted to say 10 and you would be correct. However, by the logic explained above 10¹( Modulo 11) is also -1.

Also, 10² is, simply 10¹.10¹. So, 10²( Modulo 11) is 10¹( Modulo 11).10¹( Modulo 11). We say that the Modulo operation is distributive over multiplication but we will discuss that in another post soon.

So, 10²( Modulo 11) is (-1)² = 1. Similarly 10³( Modulo 11) is (-1)³ = -1 and subsequent powers of 10 alternate between 1 and -1.

It is now time to put all this together. Let us take another look at Expression 2 from earlier.

A.(-1⁷).(10⁷) + B.(-1⁶).(10⁶) + C.(-1⁵).(10⁵) + D.(-1⁴).(10⁴) + E.(-1³).(10³) + F.(-1²).(10²) + G.(-1¹).(10¹) + H.(-1⁰).(10⁰)

Now we know where the (-1)ⁿ in each term comes from. It is just the remainder when (10)ⁿ is divided by 11. Knowing this we can express the ( Modulo 11) operation on the number ABCDEFGH as follows:

A.(-1) + B.(+1) + C.(-1) + D.(+1) + E.(-1) + F.(+1) + G.(-1) + H.(+1)

which further reduces to

– A + B – C + D – E + F – G + H (Expression 3)

This is a neat re-arrangement of the original Test for Divisibility by 11 which reads as follows:

(B + D + F + H) – (A + C + E + G) (Expression 4)

If this expression equals 0 or a multiple of 11 (we can contract this to say, “If this expression equals an integral multiple of 11”) then the original number, ABCDEFGH, is divisble by 11.

Expression 3 is interesting because it says that we need not add up the even place digits and the odd place digits separately but that we can add and subtract them from left-to-right (or right-to-left). Here is the shortcut way:

(A – B) + (C – D) + (E – F) + (G – H) (Expression 5)

Granted this expression yields a value opposite in sign to Expression 4 but we are concerned only whether the result is a multiple of 11; the sign does not matter.

Let us take an example at this point. Take the number 8492. Here are the steps to applying our shortcut:

• Subtract the right from the left of the first set of two digits and hold on to this result
• In our case that gives 8 – 4 = 4
• Do this for the next set of two digits and keep doing this till all digits are covered and keep adding the result – keeping in mind the sign of a subtraction – to the first result
• For us the next subtraction is 9 – 2 = 7. Adding the 7 to the 4 we had earlier gives us 11
• Examine the final result. If it is an integral multiple of 11 (…. +22, +11, 0, -11, -22 ….) then the original number is divisble by 11. Otherwise it is not.
• In our case the sum comes out to be 11. Therefore, 8492 is divisible by 11.

What we have ignored in this method is the case of numbers with odd total digits; 5-digit numbers, for instance. In our quest for speed we started from the left but did not care to check if the last set of digits were, indeed, a pair.

The problem is easily fixed, however. If you get a last digit that doesn’t have anything on its right simply treat the imaginary digit on the right as a Zero. This will not affect the divisibility by 11 (try and figure out why this is).

So, for a number like 15092 the steps would be as follows:

• 1 – 5 = -4.
• 0 – 9 = -9. Adding this to -4 we get -13
• 2 – 0 = 2. Adding this to -13 we get -11
• Since -11 is an integral multiple of 11, 15092 is divisble by 11

Now would be a good time to revise these concepts. Because after this, we will try and solve the question that we started this exercise with.

So, is 411281013 divisible by 11?

Let’s check:

1. 4 – 1 = 3.
2. 1 – 2 = -1. Adding this to 3 we get 2
3. 8 – 1 = 7. Adding this to 2 we get 9
4. 0 – 1 = -1. Adding this to 9 we get 8
5. 3 (- 0) = 3. Adding this to 8 we get 11

What can you say about the divisibility of 411281013 by 11?

Next Steps

You have picked up some important skills through this exercise. We recommend that you polish these skills by practising often. You can extend your abilities by trying the following:

• Use the same procedure to understand the divisibility tests for other numbers. The test for 9 is the easiest. Prime numbers are harder but completely decipherable.
• We have established a method that works for all integers. However, for numbers of fixed length – say, 4-digit licence plate numbers in India – several tweaks can be made to this method to speed it up even further. We urge you to try and figure out some of these methods for yourself.
• Notice how 10ⁿ( Modulo 11) alternates between +1 and -1 for different values of n? This behaviour – of repetition – is called cyclicity. We say that 11 has a cyclicity of 2 with respect to the Modulo function (because the result repeats every 2 powers of n). Numbers like 7, 13 and others will have larger cyclicities so you may need to persevere longer to see patterns emerging.
• Cyclicity is a fundamental behaviour in Mathematics and has a number of uses. We recommend you research this topic more. Specifically, look for cyclicity in division and in calculating the last-n digits of an exponent (17³⁷, for instance)

Stay tuned! For all the strategy tips for cracking your IIT-JEE exams . We will visit several more such concepts in this blog periodically.

If you liked what you read , please spare a moment to share your feedback, ideas and tips in the comments below. 

All the best!

Team Meritnation