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Saturday, October 13, 2012

Digital Roots Revisited: When Not To Get Carried Away With Them

In this earlier post, I talked about three properties of digital roots.  In simple mathematical terms, if the digital root of a number a with respect to base n is expressed as Dn(a), I provided a proof of the following properties of digital roots:
  • Dn(a + b) = Dn(Dn(a) + Dn(b))
  • Dn(a - b) = Dn(Dn(a) - Dn(b))
  • Dn(a * b) = Dn(Dn(a) * Dn(b))
 While working with my children on some brain-teasers, I thought I had come across another property of digital roots that I thought was quite interesting.  To set up the background, the problem I posed to my children as as follows:

Find the digital root with respect to 9 (i.e. the traditional definition of digital roots) of 1777^1777 (where ^ denotes the operation of raising to the power of).  I was not sure my children knew the properties of digital roots, but not only did they know the three above, but they used another one which I was not sure was even true.

The way I would solve the problem I posed, I would calculate D9(1777) as 4.  I know that the digital root of 1777*1777 is 7 because that is the digital root of 4*4, which is 16.  I also know that the digital root of 1777^3 would be 1 (the digital root of both 7*4 or 4*4*4 is 1).  The digital root of 1777^4 would be 4 (the digital root of 1*4 or 4*4*4*4 are both 4).  So, the pattern of digital roots as we raise 1777 to different powers is 4, 7, 1, 4, ... .  The pattern has a period of 3, so I would divide 1777 (the exponent) by 3 to get a remainder of 1.  This would then point me to the first number in my series of digital roots (4) as the answer to my problem.

My children, though, got the better of me.  They arrived at this answer well before I could.  How did they do it?  They took the digital root of both 1777's, and decided to find the digital root of 4^4 as the solution to my problem.  That gave them D9(256), which is 4.

So, is it true that D(a^b) = D(D(a)^D(b))?  Did I miss an important property of digital roots that can make some problems simpler than the method I was using for them all along?  I decided to verify if I could prove the correctness of what my children did, and this post is the result of that effort.

I started working out the proof using n as the base of my digital root for the sake of generality.  If that works, then we can be assured that it will work for the traditional definition of digital root, which uses 9 as the base.  But the general nature of the proof will give us confidence that it can be used with other bases also.

Let a = nw + x
Let b = ny + z

So, Dn(a) = x, and Dn(b) = z.

Let a^b = c.  Thus c = (nw+x)^(ny+z).

Now, we know that p^(q+r) = p^q * p^r.

So, we can write c as (nw+x)^ny * (nw+x)^z.  This is a product of two terms.  We can use the application of digital roots to multiplication to say that Dn(c) = Dn(Dn((nw+x)^ny)*Dn((nw+x)^z)).  Don't let the complexity of the expression throw you off.  It is basically a rewriting of D(a * b) = D(D(a) * D(b)), with a being equal to (nw+x)^ny and b being equal to (nw+x)^z.

Now, let us look at the number (nw+x)^ny.  This is basically (nw+x)*(nw+x)*(nw+x)*..., ny times.  When you expand out this product, you will find that all the terms in the answer except x^ny contain nw in them, and are therefore divisible by n without a remainder.  So, they don't contribute to the digital root at all.  So, Dn((nw+x)^ny) = Dn(x^ny).  By a similar logic, we can easily see that Dn((nw+x)^z) = Dn(x^z).

So, Dn(c) = Dn(Dn(x^ny) * Dn(x^z)).

Now, for my proof to work, I need to get Dn(x^ny) to be equal to 1 so that Dn(c) = Dn(Dn(x^z)).  I actually got stuck at this point.  And it was not from lack of trying.  I tried everything possible to try to prove that Dn(x^ny) = 1.  I was so brainwashed into thinking that the property was true, and it was my ability to prove it that was lacking that I wasted a lot of time mulling over the problem.  Basically, given the generality of x, n and y, this would mean that when you raise any number to a multiple of the digital root base, the digital root of the result should be equal to 1.

This is certainly true for some numbers like 4 (the digital roots with respect to 9 of whose powers cycle through 1 at multiples of 3).  But it is not true for numbers such as 2 (2^9 = 512, whose digital root is 8), 5 (5^9 = 1,953,125, whose digital root is 8) or 8 (8^9 = 134,217,728, whose digital root is 8).

I then used this knowledge to come up with a simple problem which actually disproves the property.  If you try to use this for 2^18 using 9 as a base for your digital roots, you will immediately see that the property does not work.  The digital root of 2 is 2 and that of 18 is 9.  So, the digital root of 2^18 should be the same as the digital root of 2^9 if this property were true.  But, unfortunately, the digital root of 2^9 is 8 and that of 2^18 is 1.

In fact, it turns out that when you use 9 as your base for digital roots, the property is true only when the digital root of the number being raised to powers (a in our case) is either 1, 4, 7 or 9.  It does not work for any other numbers.  You can verify that it does not work by calculating the digital roots of 2^10, 3^10, 5^10, 6^10 and 8^10 (if this property were true, the digital roots of these numbers should be 2, 3, 5, 6 and 8 respectively, and you can easily verify that they are not).

So, my children had gotten lucky because I asked them for 1777^1777, and 1777 has a digital root of 4.  And 4 just happens to be one of the few numbers for which the property actually works (at least as far as digital roots with respect to 9 go.  It does not work even for 4 when you work with digital roots with respect to other bases such as 7, for instance).

So, when you are asked to work out the digital root of a^b, where a and b are large numbers, you have to work it out the traditional way by figuring out the series of digital roots of powers of the digital root of a, find their period, divide b by that period to find the remainder, and then figure out what digital root that corresponds to.  You cannot short-circuit the process by taking the digital roots of a and b, and using them in an exponentiation operation to come up with the answer.  Yes, it will work if the digital root of a is 4 and you are working with a base of 9, but won't work with a lot of other numbers!

What is the point of all this then?  The point is that, sometimes, things work because they are special cases, not because they are true in general.  When your first encounter with something is that special case, you may be tempted to conclude that it works all the time and draw a generalization from it.  The point of this post is that it pays to be cautious.  If possible, work the math out and convince yourself with a proof that the case you encountered is not special in some way.  Don't get carried away with a generalization that is not justified.  Good luck!

8 comments:

Aniket Kulkarni said...

Finding it useful! Keep bestowing knowledge- a math boy

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