*One cannot escape the feeling that these mathematical formulas have an independent existence and an intelligence of their own, that they are wiser than we are, wiser even than their discoverers, that we get more out of them than was originally put into them.***Heinrich Hertz**

Algebra, often perceived as a daunting challenge, possesses incredible potential to solve complex real-world problems. In this blog post, we’ll explore the fascinating world of algebra, focusing on its ability to unveil hidden digits. When confronted with the task of uncovering missing digits, algebra equips us with the means to create precise equations based on given data, enabling us to effortlessly determine the unknown variables.

This is the second part of the blog post titled **The Power of Algebra in Finding the Missing Digit**. The **first part** is available here, which presents another variation of this topic. Join us on this mathematical exploration as we tackle the problem at hand.

- Ask your friend,
**Number**, to select a multi-digit number. Suppose the chosen number is**9478**. **Number**then rearranges the digits of the number to form another number, 9874.- Next,
**Number**subtracts this number from the original one, resulting in 9874 – 9478 = 396. - Now, ask
**Number**to cross out any one of the three digits; let’s say**Number**crosses out 3.**Number**will then reveal the remaining digits, which are 9 and 6 in this case. - Add these remaining digits provided by your friend,
**Number**, and get 9 + 6 = 15. - 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 + 6 = 15 here).
- Therefore, the crossed-out digit is 18 – 15 = 3.

Clearly, this statement holds true for the number 9478. Does it apply to any four-digit number? Absolutely! The proof for this is provided below. Does it also hold for a three-digit number? Yes, it does. Now, what about a five-digit number? Let’s explore if this principle is applicable to a six-digit number as well.

- Consider the four-digit number 194856.
- Rearrange the digits to form another number: 986541.
- Subtracting the original number 194856 from 986541 results in 986541 – 194856 = 791685.
- Now, let’s cross out the digit 9.
- The sum of the remaining digits is 7 + 1 + 6 + 8 + 5 = 27.
- The nearest multiple of 9 is 27, and the difference between 27 and 27 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 number obtained after rearranging its digits, the difference will always be divisible by 9.

Here is the proof for a four-digit number: consider *xyzt* as a four-digit number, where *t* is the units digit, *z* is the tens digit, *y* is the hundreds digit, and *x* is the thousands digit. We can express this number as:

** xyzt = 1000x + 100y + 10z + t**.

When we rearrange the digits of this number and form new four-digit numbers, there are 4! = 24 possible arrangements:

*xyzt*,*xytz*,*xzty*,*xzyt*,*xtyz*,*xtzy*,*yxzt*,*yxtz*,*ytxz*,*ytzx*,*yztx*,*yzxt*,*zxty*,*zxyt*,*zyxt*,*zytx*,*ztxy*,*ztyx*,*txyz*,*txzy*,*tzxy*,*tzyx*,*tyxz*,*tyzx*.

We now consider a number after rearranging the digits of the number *xyzt*:

** txyz = 1000t + 100y + 10z + x**.

A simple calculation shows that

xyzt–txyz= (1000

x+ 100y+ 10z+t) – (1000t+ 100x+ 10y+z)= 900

x+ 90y+ 9z– 999t= 9(100

x+ 10y+z– 111t).

But 9(100*x* + 10*y* + *z* – 111*t*) is, of course, divisible by 9. Therefore, when we subtract from a number the rearranged number, the result is always divisible by 9.

**Case 1**. Let us assume that 9(100*x* + 10*y* + *z* – 111*t*) is a four-digit number. Then

9(11*x* + *y*) = *abcd* = 1000*a* + 100*b* + 10*c* + *d*.

Recall that the three-digit number *abcd* is divisible by 9, so that (*a* + *b* + *c* + *d*) is also divisible by 9. Since *a*, *b*, *c* and *d* are digits, maximum value of each of them is 9.

Let us cross out the digit *b*. Essentially, the sum (*a* + *c* + *d*) will be 9, 18 or 27 only. It follows that *b* = 9 – (*a* + *c* + *d*) or 18 – (*a* + *c* + *d*) or 27 – (*a* + *c* + *d*) and the result follows.

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

**Case 3**. Let us assume that (11*x* + *y*) is a two-digit number. Proceed as above to get the result.

**Case 4**. Let us assume that (11*x* + *y*) is a single-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 nine-digit number 987654321.
- Rearrange the digits and form a new number 123456789.
- Their difference is 987654321 – 123456789 = 864197532.
- Let’s cross out the digit 8.
- The sum of the remaining digits is 6 + 4 + 1 + 9 + 7 + 5 + 3 + 2 = 37.
- The nearest multiple of 9 is 45, and the difference between 45 and 37 is 8.
- Therefore, the crossed-out digit is 8.

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