 # The Power of Algebra in Finding the Missing Digit – Part 1

In mathematics he was greater than Tycho Brahe or Erra Pater.
For he, by geometric scale, could take the size of pots of ale.
Resolve, by sines and tangents straight, if bread or butter wanted weight;
And wisely tell what hour of the day the clock does strike, by Algebra.

Samuel Butler

Algebra is a branch of mathematics that deals with symbols and the rules for manipulating these symbols to solve equations and understand relationships between variables. It helps us analyze and solve problems by representing unknown quantities with variables and then using equations to find their values. Algebra, often regarded as a daunting subject by many, is a powerful tool that can unravel a myriad of real-world problems. In this blog post, we will delve into the remarkable capabilities of algebra, specifically in the context of discovering elusive digits. When faced with the challenge of finding missing digits, algebra empowers us to construct equations grounded in provided information and effortlessly solve for the unknown variables. Let’s embark on this mathematical journey and consider the problem at hand.

• Suppose the chosen number is 947.
• Then, Digit adds up the individual digits of the selected number, which results in 9 + 4 + 7 = 20.
• Digit then subtracts this total from the original number, resulting in 947 – 20 = 927.
• Next, ask Digit to cross out any one of the three digits; let’s say Digit crosses out 2.
• Now, Digit will reveal the remaining digits, which are 9 and 7 in this case.
• You add these remaining digits provided by your friend Digit and get 9 + 7 = 16.
• Then, find the nearest multiple of 9 to the obtained sum, which, in this case, is 18.
• The missing digit is the difference between the multiple of 9 (18 in this case) and the sum of the known digits (9 + 7 = 16 here). Therefore, the crossed-out digit is 18 – 16 = 2.

Clearly, this statement holds true for the number 947. Does it apply to any three-digit number? Absolutely! The proof for this is provided below. Does it also hold for a two-digit number? Yes, it does. Now, what about a four-digit number? Let’s explore if this principle is applicable to a four-digit number as well.

• Consider the four-digit number 1948.
• The sum of its digits is 1 + 9 + 4 + 8 = 22.
• Subtracting 22 from the original number yields 1948 – 22 = 1926.
• Let’s cross out the digit 9.
• The sum of the remaining digits is 1 + 2 + 6 = 9.
• The nearest multiple of 9 is 9, and the difference between 9 and 9 is 0.
• Therefore, the crossed-out digit is either 0 or 9.

Note. When the difference is 0, the possibilities are 0 and 9 only.

What’s the magic? No matter what the number is, if you subtract from it the total number of its digits, the balance will always be divisible by 9.

Here is the proof for a three-digit number: consider xyz as a three-digit number, where z is the units digit, y is the tens digit, and x is the hundreds digit. We can thus write

xyz = 100x + 10y + z.

From this number, we subtract the sum of its digits x + y + z and obtain:

But 9 (11x + y) is, of course, divisible by 9. Therefore, when we subtract from a number the sum of its digits, the result is always divisible by 9.

Clearly, (11x + y) is a two or three-digit number.

Case 1. Let us assume that (11x + y) is a three-digit number. Then

9(11x + y) = abc = 100a + 10b + c.

Let us cross out the digit b. Recall that the three-digit number abc is divisible by 9, so that (a + b + c) is also divisible by 9. Since a, b, and c are digits, the maximum value of each of them is 9.

Let’s cancel the middle-digit b. Essentially, the sum (a + c) will be 9 or 18 only. It follows that b = 9 – (a + c) or 18 – (a + c) and the result follows.

Case 2. Let us assume that (11x + y) is a two-digit number. Proceed as above to get the result.

It may happen that the sum of the digits you are told is divisible by 9 (for example, 4 and 5). That shows that the digit your friend has crossed out is either 0 or 9, and in that case, you have to say that the missing digit is either 0 or 9.

I am sure that, you can produce a general proof for this.

• Finally, consider the eight-digit number 12345678.
• The sum of its digits is 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 = 26.
• Subtracting 26 from the original number yields 12345678 – 22 = 12345656.
• Let’s cross out the digit 1.
• The sum of the remaining digits is 2 + 3 + 4 + 5 + 6 + 7 + 8 = 35.
• The nearest multiple of 9 is 36, and the difference between 36 and 35 is 1.
• Therefore, the crossed-out digit is 1.

This is the first part of the post showcasing the power of algebra. A second part will be posted in a few days with a slight variation. Please do read that post as well.

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