Perhaps I could best describe my experience of doing mathematics in terms of entering a dark mansion. You go into the first room and it’s dark, completely dark. You stumble around, bumping into the furniture. Gradually, you learn where each piece of furniture is. And finally, after six months or so, you find the light switch and turn it on. Suddenly, it’s all illuminated and you can see exactly where you were. Then you enter the next dark room.

**Andrew Wiles**

Glad you came by. I wanted to let you know I appreciate your spending time here at the blog very much. I do appreciate your taking time out of your busy schedule to check out **Math1089**!

Consider any two-digit number. We can write them as below:

- 25 = 10 × 2 + 5
- 47 = 10 × 4 + 7
- 69 = 10 × 6 + 9
- 81 = 10 × 8 + 1

In general, any two-digit number *ab *made of digits *a* (ten’s)and *b* unit)can be written as *ab *= 10 × *a *+ *b *= 10*a *+ *b.* When the digits get reversed, new number look like *ba *= 10 × *b *+ *a *= 10*b *+ *a.* We can impose an extra condition, like *a* > *b* or *a* < *b* or *a* = *b* here.

Fact 1. Consider a two-digit number. Now reverse the digits (satisfy yourself that the digits remain same). Add the two numbers so obtained. ThisSUMis divisible by11and thesumof the digits.

Initially to verify, let us choose 31 as the number (one may take a different number also). Then,sumof the digits = 3 + 1 = 4. After reversing the digits, new number will be 13. ThenSUM= 31 + 13 =44and it is divisible by4and11.

Fact 2. Consider a two-digit number. Now reverse the digits. Subtract the numbers (smaller from the larger one) so obtained. ThisDIFFERENCEis divisible by3,9and thedifferenceof the digits. Note that, thisDIFFERENCEmay be zero.Initially to verify, let us choose 59 as the number. Then,

differenceof the digits= 9 – 5 =4. After reversing the digits, new number will be 85. ThenDIFFERENCE= 95 – 59 =36and it is divisible by3,9and4.

**Proof**. Let *ab *= 10 × *a *+ *b *= 10*a *+ *b* be the given two-digit number. The new number after reversing the digits will be *ba *= 10 × *b *+ *a *= 10*b *+ *a.* Then

(**i**) **SUM** of the numbers = *ab* + *ba* = (10*a *+ *b*) + (10*b *+ *a*) = 11(*a *+ *b*). Clearly, this is divisible by 11 and the **sum** *of the digits*.

(**ii**) **DIFFERENCE** of the numbers

=

ab−ba= (10a+b) − (10b+a) = 9(a−b), ifab>ba;=

ba−ab= (10b+a) − (10b+a) = 9(b−a), ifab<ba;

Clearly, this is divisible by 3, 9 and the **difference** *of the digits*.

Consider any three-digit number. We can write them as below:

- 325 = 100 × 3 + 10 × 2 + 5
- 647 = 100 × 6 + 10 × 4 + 7
- 869 = 100 × 8 + 10 × 6 + 9
- 580 = 100 × 5 + 10 × 8 + 5

In general, any three-digit number *abc *made of digits *a* (hundred’s), *b* (ten’s)and *c* (unit) can be written as *abc *= 100 × *a* + 10 × *b* + *c* = 100*a* + 10*b* + *c.* In the same way, we can write *cab *= 100*c *+ 10*a *+ *b* and *bca *= 100*b *+ 10*c *+ *a*.

Consider the three-digit number *abc*. Form a new number with the *one’s digit shifted* to the *left end* of the number, so the number is *cab*. Repeat the procedure and form the number *bca*. We get *cba* from the number *abc*, after *reversing the digits*. We can impose one or more extra conditions, like *a* > *c* or *a* < *b* or *c* = *b* etc.

Fact 3. Consider a three-digit numberabc. Now form the numbers likecabandbcawith theone’s digit shiftedto theleft endof the number. Add the three numbers so obtained. ThisSUMis divisible by3,37,111and thesumof the digits.Initially to verify, let us choose 943 as the number (one may take a different number also). Then,

sumof the digits= 9 + 4 + 3 =16. Shifting the digits to theleft endof the number, we can form 394 and 439. Now,SUMof all the numbers = 943 + 394 + 439 =1776and it is divisible by3,37,111and thesumof the digits.

Fact 4. Consider a three-digit number. Now reverse the order of the digits. Subtract the numbers so obtained. ThisDIFFERENCEis divisible by3,9,11,33,99and thedifferenceof the extreme digits. Note that, thisDIFFERENCEmay be zero.To verify, let us choose 539 as the number. Then,

differenceof the extreme digits= 9 – 5 =4. After reversing the order of the digits, new number will be 935. ThenDIFFERENCE= 935 – 539 =396and it is divisible by3,4,9,11,33and99.

**Proof**. Let *abc *= 100*a *+ 10*b *+ *c* be the given three-digit number. Shifting the digits, we can form *cab *= 100*c *+ 10*a *+ *b* and *bca *= 100*b *+ 10*c *+ *a*. Then

(**iii**) **SUM** of the numbers *abc *+ *cab *+ *bca*

= (100*a *+ 10*b *+ *c*) + (100*c *+ 10*a *+ *b*) + (100*b *+ 10*c *+ *a*)

= 111(*a *+ *b *+ *c*) = 3 × 37 × (*a *+ *b *+ *c*),

which is divisible by **3**, **37**, **111** and the **sum ***of the digits*.

(**iv**) **DIFFERENCE** of the numbers

=

abc−cba= (100

a+ 10b+c) – (100c+ 10b+a)= 100

a+ 10b+c– 100c –10b–a= 99

a– 99c= 99(a–c), ifa>c.

If *c *> *a, *then the **DIFFERENCE** of the numbers

=

cba−abc= (100

c+ 10b+a) – (100a+ 10b+c)= 99

c– 99a= 99(c–a).

If the digits are equal, of course the difference is 0.

In any case, the **DIFFERENCE** is divisible by **3**, **9**, **11**, **33**, **99** and the **difference ***of the extreme digits*.

(**a**) Consider a two-digit number *ab *and the number by reversing its digits, i.e., *ba*. Consider the sum *ab *+ *ba *and let the sum be a three-digit number *dad*.

- Why the sum
*a*+*b*cannot exceed 18? - Is
*dad*a multiple of 11? - What can be the maximum value of
*dad*? - Find the values of
*a*and*d*.

(**b**) You have seen that a number 540 is divisible by 10. It is also divisible by 2 and 5 which are factors of 10. Similarly, a number 144 is divisible 9. It is also divisible by 3 which is a factor of 9. Can you say that if a number is divisible by any number *n*, then it will also be divisible by each of the factors of *n*?

(**c**) Consider a three-digit number *abc *= 100*a *+ 10*b *+ *c* = 11(9*a *+ *b*) + (*a *– *b *+ *c*). If the number *abc *is divisible by 11, then what can you say about (*a *– *b *+ *c*)? Is it necessary that (*a *+ *c *– *b*) should be divisible by 11?

(**d**) Consider a four-digit number *abcd *= 1000*a *+ 100*b *+ 10*c *+ *d* = 11(91*a *+ 9*b *+ *c*) + [(*b *+ *d*) – (*a *+ *c*)]. If the number *abcd *is divisible by 11, then what can you say about [(*b *+ *d*) – (*a *+ *c*)]?

(**e**) From (**c**) and (**d**) above, can we say that a number will be divisible by 11 if the difference between the sum of digits at its odd places and that of digits at the even places is divisible by 11?

Your suggestions are eagerly and respectfully welcome! See you soon with a new mathematics blog that you and I call **“****Math1089 – Mathematics for All!**“.