A list of numbers starts with a two-digit number. The second number in the list is derived by adding the sum of the digits of this first number to the first number itself. The result is another 2-digit number. The third number in the list is then formed by adding the sum of the digits of the second number to the second number. The third number in the list is 44. What is the first number in the list?

At first glance, it seems pretty straightforward. In fact, when my daughter read it to me, my first thought was that the solution would end up being a system of simultaneous equations from which the answer would pop out. In fact, since my daughter is learning about simultaneous equations right now, that seemed a pretty safe guess. But I was quite mistaken!

I immediately started out formulating the problem as below:

Let 10x + y be the first number in the list. The second number in the list is, therefore, 10x + y + x + y. This can be simplified into 11x + 2y. Now, 11x + 2y is equal to a two-digit number also. Let us represent that number by 10a + b. Now, the third number in the series will be 10a + b + a + b. Thus, 11a + 2b. And it is given that 11a + 2b = 44.

Excellent so far. But looking closely at the system above, it is obvious that there are only 2 equations in 4 unknowns:

11x + 2y = 10a + b

11a + 2b = 44

Obviously, this system is not solvable by algebraic methods. And, it is unlikely to be the solution methodology envisioned by my daughter's teacher since her class is barely into solving simultaneous equations in 2 variables. I don't expect even my older daughter to solve a system of 4 simultaneous equations in 4 variables (assuming I would find two more equations in a, b, x and y hiding out there somewhere, waiting to be discovered). And to be perfectly honest, I don't look forward to solving such a system by hand either!

Obviously, I am missing something. But what? I can't think of anything intuitive that I have missed in the above analysis. But I must be.

By the way, it is easy enough to solve this problem by trial and error. Given that 11a + 2b = 44, and 0<=a<=9 and 0<=b<=9, it is easy that see that a can not be 1 or 2 (then b would have to be greater than 9 and therefore cannot be the digit of a number), and a can not be 3 because 33 + an even number (which is what 2*b is) can not be 44.

So, the second number is 40, and applying the logic backwards one more step with x and y, you get the first number in the list as 29. So, the list is 29, 40, 44, ...

But, being able to solve a problem by trial and error is never any fun. It never gives you the sense of satisfaction you get when you solve the problem correctly so that you derive a general formula for the answer that you can apply without having to stumble around in the dark. What if I had been given the 11th number in the list and asked to derive the first number of the list? That would be a lot of trial and error to work through to get to the final answer.

I must admit that I hate problems that involve the digits of a number. The basic problem is that the digits of a number are not at all obvious when the number is expressed as a polynomial unless the coefficients are all powers of 10.

Take the second number of this list, for instance. It can be expressed as 11x + 2y. But what exactly are the digits of this number? Is there a formula expressible in x and y that will tell me what the digits of the number are? In this case, 11x + 2y turned out to be the number 40. 4 is not even divisible by or a factor of 11. Similarly, 11a + 2b turned out to be 44. But in the final answer, b was not even equal to 2!

I guess the puzzle this time is: Is there a better method than guess and check to solve problems like this? What am I missing that is causing me not to be able to solve this puzzle (and others similar to this) algebraically?

Now, should I be disturbed by my inability to solve this problem algebraically given that I was able to solve it by trial and error? I think so. Let me pose a slightly different problem for you. A list starts with a certain two-digit number. The second number is list is derived by adding to that number 4 times its first digit (ten's digit) and subtracting 5 times the second digit (units digit). The result is another two-digit number. The third number is derived by adding to the second number 4 times the second number's tens digit and subtracting 5 times the second number's units digit. The process is continued on and on. The sixth number in the list is 82. What is the first number in the list?

You see the problem with the trial and error problem-solving technique now? What if I made the list start with a 3 or 4-digit number, and derived the subsequent numbers in the list using even more complicated mathematical manipulations? And then gave you the 75th number of the list and asked you for the 1st number? Or even gave you the 1st number and asked you for the 75th? Without an algebraic way of going back and forth down the list quickly, it would become pretty painful pretty quickly.

Now, I do understand that these types of problems are rather unique in one respect: Each of these problems is written based on using a particular base for the numbers (in this case, base 10), and can be solved only in that base. In most other mathematical problems, the base in which numbers are expressed is completely irrelevant. But any time the digits of a number enter the equation, the base in which the numbers are expressed becomes extremely important.

So, where do you think my mental block is? Is guess and check the best method for problems of this sort? Is there a better method I am not thinking of? Are problems which do not have unique solutions without reference to a particular base of operations doomed to trial and error methods of solution? How do I take the base of operations into consideration explicitly in that case when trying to derive an algebraic solution? Anybody willing to help me in this effort? Please feel free to chime in with comments on this post. Good luck!

## 8 comments:

Dear Webmaster,

I don't know if what follows could help you...but let's try something.

If 11a + 2b = 44 then, 10a + a + 2b = 44. But, as 11,4. So a>2 and a<10.

Let's say that a = 2+a' wuth a'<8.

From 11a+2b = 44, we reaches 22+11a'+2b = 44 so 11a'+2b = 22 from where we concludes that if a'=2,b=0 but if a=1 b can't be equal to 5,5. Then a'=2 and b=0 so a = 4 and b = 0.

I seems very closed to the "try and errors method" but the equation 11a'+2b=22 seems more accurate than 11a+2b=44.

I hope this helped you, in one way or another...

Best regards,

D. Auroy (you've got my mail address in your mailbox).

Dear Webmaster,

I don't know if what follows could help you...but let's try something.

If 11a + 2b = 44 then, 10a + a + 2b = 44.

But, as 11,4.

So a>2 and a<10.

Let's say that a = 2+a' with a'<8.

From 11a+2b = 44,

we reaches

22+11a'+2b = 44

so 11a'+2b = 22

from where we concludes that if

a'=2,b=0 but if a=1 b can't be equal to 5,5.

Then a'=2 and b=0 so a = 4 and b = 0.

I seems very closed to the "try and errors method" but the equation 11a'+2b=22 seems more accurate than 11a+2b=44.

I hope this helped you, in one way or another...

Best regards,

D. Auroy (you've got my mail address in your mailbox).

PS : I hope the readers could read this new message in a better way.

I sent you a mail with the "good" message. Please don't take attention to the two first.

I read your comment with interest, and reducing 11a + 2b = 44 to 11a' + 2b = 22 seems to help, but I am not able to follow the logic as to how you accomplished it. In particular, in your comments, you have the line "But, as 11,4." I don't know what you meant to write, but that sentence must be important to the further derivation in your method. Please post back with more details. Thank you.

It's very difficult to write an explanation on your blog : the comments box always erase a big part of my answer...

I tried to send you a mail with the global answer...

I hope you get it.

Dear Annath,

With just one equation and 4 digits to be found, there are bound to be many solutions, mathematically, not necessarily in reality. Since digits can't take other than integer values, some trial and error approach is anyway required. In my opinion, if we solve the last equation then we can find the 2nd last number and then the original number could be found. It's virtually impossible to solve directly. If we perform 75 iterations then and just reveal the last number, it won't be possible to find the starting number unless we come backwards one by one.

Just my 2 cents.

Thanks.

AB

You are right. I wrestled with it for several days, and finally came to the conclusion that perhaps there are problems like this that are impossible to solve cleanly. It still makes me uncomfortable, but sometimes, you have to move on rather than obsess over one particularly difficult problem.

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