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
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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 = 10a + b. When the digits get reversed, new number look like ba = 10 × b + a = 10b + 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. This SUM is divisible by 11 and the sum of the digits.
Initially to verify, let us choose 31 as the number (one may take a different number also). Then, sum of the digits = 3 + 1 = 4. After reversing the digits, new number will be 13. Then SUM = 31 + 13 = 44 and it is divisible by 4 and 11.
Fact 2. Consider a two-digit number. Now reverse the digits. Subtract the numbers (smaller from the larger one) so obtained. This DIFFERENCE is divisible by 3, 9 and the difference of the digits. Note that, this DIFFERENCE may be zero.
Initially to verify, let us choose 59 as the number. Then, difference of the digits = 9 – 5 = 4. After reversing the digits, new number will be 85. Then DIFFERENCE = 95 – 59 = 36 and it is divisible by 3, 9 and 4.
Proof. Let ab = 10 × a + b = 10a + b be the given two-digit number. The new number after reversing the digits will be ba = 10 × b + a = 10b + a. Then
(i) SUM of the numbers = ab + ba = (10a + b) + (10b + 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), if ab > ba;
= ba − ab = (10b + a) − (10b + a) = 9(b − a), if ab < 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 = 100a + 10b + c. In the same way, we can write cab = 100c + 10a + b and bca = 100b + 10c + 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 number abc. Now form the numbers like cab and bca with the one’s digit shifted to the left end of the number. Add the three numbers so obtained. This SUM is divisible by 3, 37, 111 and the sum of the digits.
Initially to verify, let us choose 943 as the number (one may take a different number also). Then, sum of the digits = 9 + 4 + 3 = 16. Shifting the digits to the left end of the number, we can form 394 and 439. Now, SUM of all the numbers = 943 + 394 + 439 = 1776 and it is divisible by 3, 37, 111 and the sum of the digits.
Fact 4. Consider a three-digit number. Now reverse the order of the digits. Subtract the numbers so obtained. This DIFFERENCE is divisible by 3, 9, 11, 33, 99 and the difference of the extreme digits. Note that, this DIFFERENCE may be zero.
To verify, let us choose 539 as the number. Then, difference of the extreme digits = 9 – 5 = 4. After reversing the order of the digits, new number will be 935. Then DIFFERENCE = 935 – 539 = 396 and it is divisible by 3, 4, 9, 11, 33 and 99.
Proof. Let abc = 100a + 10b + c be the given three-digit number. Shifting the digits, we can form cab = 100c + 10a + b and bca = 100b + 10c + a. Then
(iii) SUM of the numbers abc + cab + bca
= (100a + 10b + c) + (100c + 10a + b) + (100b + 10c + 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
= (100a + 10b + c) – (100c + 10b + a)
= 100a + 10b + c – 100c – 10b – a
= 99a – 99c = 99(a – c), if a > c.
If c > a, then the DIFFERENCE of the numbers
= cba − abc
= (100c + 10b + a) – (100a + 10b + c)
= 99c – 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 = 100a + 10b + c = 11(9a + 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 = 1000a + 100b + 10c + d = 11(91a + 9b + 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!“.