Please take a look at the Fun With Num3ers blog by Benjamin Vitale (Twitter: @BenVitale). This is a fun blog working with number theory. It is also an inspiration for me to blog. Thanks, Ben!

Eddie

Link: http://benvitalenum3ers.wordpress.com/

On to today's blog entry!

Two Digit Numbers

Let AB represent a two digit number, where A and B each represent a digit 0 through 9 (excluding 00). My original goal was to find a number AB such that AB + BA = n where n contains both digits A and B. I was unsuccessful.

However, I did see something curious:

11 + 11 = 22

12 + 21 = 33

13 + 31 = 44

14 + 41 = 55

15 + 51 = 66

16 + 61 = 77

17 + 71 = 88

18 + 81 = 99

19 + 91 = 110

10 + 01 = 10 + 1 = 11

All the sums for this group are multiples of 11.

If I go on...

21 + 12 = 33

22 + 22 = 44

23 + 32 = 55

24 + 42 = 66

25 + 52 = 77

....

Once again, multiples of 11. Can we show that this is true for all sums AB + BA?

Let

*a*represent the digit A. Similarly, let

*b*represent the digit B. Then:

AB = (10a + b) and BA = (10b + a)

Then the sum AB + BA = (10a + b) + (10b + a) = 11a + 11b = 11(a + b)

Three Digit Numbers

Let's consider the number ABC. Rotate ABC one digit and add. Repeat. We get the sum ABC + BCA + CAB = n.

Let

*a*,

*b*, and

*c*represent the digits A, B, and C, respectively. Then ABC = 100a + 10b + c. We can construct two similar sums for BCA and CAB.

Then

ABC + BCA + CAB

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

= 111a + 111b + 111c

= 111(a + b + c)

The sum is ends up a multiple of 111, but not necessarily of 11. (111 is not a multiple of 11). Examples?

Example 1:

425 + 254 + 542 = 1221

1221/111 = 11

1221/11 = 111

Example 2:

963 + 639 + 396 = 1998

1998/111 = 18

1998/11 = 181 6/11

1998 is not a multiple of 11

Can we find such triples ABC, BCA, and CAB such that the sum is divisible by both 11 and 111? Hint: the least common multiple of 11 and 111 is 1,221. I have found 31 such triplets (including the trivial 0,0,0).

If you want to find the triples, have fun! I'll leave you with a few examples:

29 + 902 + 290 = 1221. (029, 902, 290)

137 + 713 + 371 = 1221. (137, 713, 371)

598 + 859 + 985 = 2442. (598, 859, 985)

254 + 425 + 542 = 1221. (254, 425, 542)

Take care and be safe!

Eddie

This blog is property of Edward Shore. 2012

Notice that when you flip the digits of a number around and take the difference between them you always get a multiple of 9:

ReplyDelete11-11 = 0

21-12 = 9

42-24 = 18

63-36 = 27

84-48 = 36

211-112 = 99

211-121 = 90

321-123 = 198 = 99*2 = 11*2*9

321-213 = 108 = 12*9

321-132 = 189 = 21*9

321-231 = 90

321-312 = 9

7546 - 4567 = 2979 = 331*9. There must be a way to prove this.

When you add a numer and it's reverse together, sometimes you get a palindrome:

11+11 = 22

21+12 = 33

65+56 = 121

123+321 = 444

Sometimes you have to do it more than once to reach a number that is a palindrome:

253+352 = 605

605+506 = 1111

143+431 = 574

574+475 = 1049

1049+9401 = 10450

10450+5401 = 15851

Somtimes you have to keep going, but you may never reach a palindrome. For example, add 89 and 98 together. Then add to that number it's reverse. Keep doing this until you reach a palindrome. I went ten or so iterations and I couldn't reach one!

Mike,

ReplyDeleteLet AB represent a two digit number, where a represents the digit A and b represents the digit B.

Then AB - BA implies that

(10a + b) - (10b + a) = 10a + b - 10b - a = 9(b - a)

Eddie

Hello Eddie !

ReplyDeleteTanks a lot for your blog, i discovered it a few days ago ans really love it !

I've done a little program for my hp35s :

LBL U

0,99901

STO A

0

STO C

RCL A

IP

STO B

XEQ U028

XEQ U026

XEQ U026

+

+

10

*

11

RMDR

X<>0?

GTO U022

1

STO+C

ISG A

GTO U006

VIEW C

RTN

LASTx

IP

10

/

FP

RTN

In several minutes (about 7) : it gives 91

30 triplets + (000) = 31

A french reader...

Pascal

Pascal,

DeleteThank you for the compliment. Nice program!

Eddie