## 1089

Posted by Chris on July 17, 2010 – 7:18 pm

A friend of mine found this:

Take a 3 digit number abc where a≠c. Reverse it. Subtract the smaller from the larger. Add the result to the reverse of the result. You get 1089. Prove it.

I really only posted it because of the previous puzzle (Reversing Numbers)

July 17th, 2010 at 9:58 pm

111 is a three digit number that will not satisfy the requirements. 111-111+111=1111089

July 17th, 2010 at 10:00 pm

Thanks Nathan. I’ll fixed it. But you should have done 111-111 = 0, then 0 + 0 = 0.

July 17th, 2010 at 10:33 pm

hmm ok so i didn’t think there was a way to do it and i’m pretty sure i was right

first off the middle number has to be 9

any middle number will = 9 during the subtraction

to get the 8 makes it so the middle number had to be 9 it can’t be 8 because there is no way to make 19 in the 1’s colom

so this means that the reversed number + the result must = 9

well you can always get the 89 but you can’t get the 10 in

actual problem using abc as the large number when reversed

abc – cba = xyz

xyz + zyx = 1089

July 17th, 2010 at 10:36 pm

sorry forgot to post where i found the info from

http://www.mathsisfun.com/1089.html

also let me clarify this

to get the 8(in the 10’s colom) makes it so the middle number had to be 9 it(the middle number) can’t be 8 because there is no way to make 19 in the 1’s colom

not sure if it was needed or not

July 18th, 2010 at 1:30 am

Take a 3 digit number abc where a≠c.

Reverse it.

Subtract the larger from the smaller.

Add the result to the reverse of the result.

You get 1089.

Prove it.

S=100*a-10*b+1*c-100*c-10*b-1*a=99*a-99*c

S=99*a-99*c=99*(a-c)

S=100*(a-c)-(a-c)

Consider 100*(a-c). If we subtract a quantity of (a-c) from it

a-c is less than or equal to 9 thus first digit is (a-c)-1.

a-c is less than or equal to 9 thus second digit must be 9

last digit is 10-(a-c) or 10-a+c

S’=reverse of S

first digit is 10-(a-c)

second digit is 9

last digit is (a-c)-1.

S+S’=100*((a-c-1)+(10-a+c))+10*(9+9)+1*((10-a+c)+(a-c-1))

S+S’=100*((a-c-1)+(10-a+c))+10*(9+9)+1*((10-a+c)+(a-c-1))

S+S’=100*(9)+10*(9+9)+1*(9)

S+S’=900+180+9=1089

However the assumptions brake down if a-c=1 as the middle digit does not exist

a-c=1 S=99 S’=99 S+S’=99+99=198 Counterexample!! e.g 645-546=99 99+99=198

So You only get 1089 if a-c is greater than 1.

Cam

July 18th, 2010 at 1:41 am

I should briefly state that abc in the calculations is chosen to be the larger of the two numbers.

Cam

July 18th, 2010 at 1:55 am

Some clarification on this part

“a-c is less than or equal to 9 thus first digit is (a-c)-1.

a-c is less than or equal to 9 thus second digit must be 9

last digit is 10-(a-c) or 10-a+c”

a-c is actually less than or equal to 8

Since a-c is less than or equal to 8:

a-c is a number between 1 and 99 thus first digit is (a-c)-1.

a-c is a number between 1 and 9 thus second digit must be 9

last digit is 10-(a-c) or 10-a+c

Cam

July 18th, 2010 at 2:53 am

Well Cam, it depends how you reverse the difference. If you always reverse it as a 3-digit number (i.e., 99 is actually 099 and its reverse is 990) then the answer is still correct even for a-c=1. The rest is OK, but I just want to point out that (at least for me) is easier to work not with 3 variables a, b, and c, but with only one, if you make the observation (you have still done it but you didn’t use the full benefit of it) that there is a one-to-one correspondence between numbers of the type a9(9-a) (these are the digits of a tr-digit number) and the absolute values of the difference of an arbitrary 3-digit number and its reverse (in the case that the first and the last digit doesn’t coincide!). Here, a is (a-c-1) in your notation and it can be between 0 and 8. Now it is straightforward that the sum a9(9-a)+(9-a)9a=1089 (again, if a=0, we reverse the number in a stranger way).

July 18th, 2010 at 5:07 am

Thanks for rescuing that slavy. I did it simply by doing the calculations using positional notation with a,b,c instead of explicit numbers, including the borrowing and carrying back. Using “,” to separate the digits.

NB a > c

a,b,c -

c,b,a

—–

a-(c+1),(10+b)-(b+1),10-(a-c) =

(a-c)-1,9,10-(a-c) then add to it’s reverse

10-(a-c),9,(a-c)-1 +

——————

9+1,8,9

= 1089

@slavy, is this significant to the other problem? (Only kidding, I think).

July 18th, 2010 at 5:46 am

Hey slavvy,

Thanks for the input.

I realized when doing that if you did 990+099=1089, but it seemed pretty hokey to me to do it that way.

If I were to ask a random person to reverse the digits of 12 I would expect 21. If they replied 21,000 ,with the justification that they could write 12 with 3 leading 0s, I would be surprised. One could justify it, but it would be a stretch.

I’m not a big fan of reducing the 3 variables down to 1 for this problem following reasons:

-have to justify/proove the 1 to 1 correspondance of the digits (not a big deal, but I’m lazy)

-easier to explain/justify solution when it is seen all the initial variables cancel out (which they almost immediately do). Less appearance of hocus pocus, and it seems more elegant (but this is, of course, subjective).

I think that if there was more steps involved in cancelling out the variables, or if the steps were complicated, I might have gone in that direction.

It just goes to show that, oddly enough, math is a discipline which has lots of room for individual style and creativity. It’s interesting to see how different people tackle a problem in different ways to get to the same result.

Cam

July 18th, 2010 at 11:55 am

@nathan, 111 can’t be an example since it should be a≠c, 111 makes a=c, that is, 1=1.

July 18th, 2010 at 2:14 pm

@ino, I added the a≠c because of and after Nathan’s post.

July 18th, 2010 at 2:45 pm

Hi, Chris! So far I don’t see connection between this problem and the previous one, but I haven’t thought much about it either. If I found something in the future, I will let you know

@Cam – I am not criticizing your solution I agree with most of your arguments (apart from proving the one-to-one correspondence, because you have basically done it and you could have just set c-a-1 to be a new variable A) and I just rewrote your solution in a less formalized way. And yes, in general reversing with a 0 in front is kind of magic, but since here we talk about 3-digit numbers from the beginning it has some justification.

July 18th, 2010 at 4:39 pm

@slavy. I only really thought that it was an amusing coincidence.

July 19th, 2010 at 7:23 am

Poorly worded question.

Take a 3 digit number abc where a≠c. Reverse it. Subtract the larger from the smaller. Add the result to the reverse of the result. You get 1089. Prove it.

The above states that if my number is 321 to reverse it and subtract the larger from the smaller. 123 – 321 = -198 and add the reverse of it.

It should read subtract smaller from largest. 321 – 123 = 198.

Add the result to the reverse of the result. 198 + 891 = 1089.

Proving it would take less than 30 minutes.

July 19th, 2010 at 7:41 am

@John24. Well spotted, my bad. I’ve fixed it for posterity. It took me about 5 minutes to solve, doing the borrows and carries did my nut in.

July 20th, 2010 at 10:02 am

This is far from a proof, but is curious nevertheless.

Note: ABC – CBA / 99 is always an integer when A≠C.

99 990

198 891

297 792

396 693

495 594

594 495

693 396

792 297

891 198

990 99

Notice all combination = 1089 when added. Also, 1089 / 99 = 11.

Suppose: ABC – CBA = DEF if FED is the reverse of DEF then FED + DEF must = 1089.

July 20th, 2010 at 2:14 pm

Using similar logic to the original question…

All 5 digit numbers have a total of 109890. Coincidentally ABCDE – EDCBA is divisible by 99. Also, 109890 / 99 = 1110.

All 7 digit numbers have a total of 10998900. Coincidentally ABCDEFG – GFEDCBA is divisible by 99. Also, 10998900 / 99 = 111100

May 9th, 2011 at 2:31 pm

I clicked on a random link and found this page. I note that John24 says that (ABC-CBA)/99 is an integer. He’s right.

ABC-CBA = 100A + 10B + C – 100C -10B – A = 100(A-C) -(A-C) = 99(A-C). If A = C then obviously get 0.

October 18th, 2012 at 2:22 pm

100(c-a)=d

10(b-b)=e

(a-c)=f

100(f+d)=900 ALWAYS

10(e+e)=180 ALWAYS (this gives it the 1000 and the 80)

(f+d)=9 ALWAYS

in total that gives you 1089

Lewis: 14

Solevd in: 30 minutes

October 18th, 2012 at 5:41 pm

Hi Lewis. You have made a bit of a mess of it. Possibly that only happened when you were writing it up, and possibly because you introduced the factors of 10 and 100 too early.

A glaring error is that you have 10(b-b) = e and that gives e = 0.

But you say 10(e+e) = 180 even though it must be 0.

I suspect that you had meant something more like as follows:

Let c:b:a – a:b:c = d:e:f – I’m using “:” to avoid confusion with multiplication.

Because c > a, when we subtract, the units part requires that we borrow from

the tens i.e. we calculate f = (10+a)-c. But then to do the tens subtraction, we

must borrow from the hundreds to find e = (10+b) – (b+1) = 9. Finally for the hundreds we calculate d = c – (a+1) and there’s no need to borrow.

So d+f = (c – (a+1)) + ((10 + a) – c) = 9 and e = 9.

Then d:e:f + f:e:d = (100d + 10e + f) + (100f + 10e + d) = 101(d+f) + 20e

= 909 + 180 = 1089.

April 17th, 2013 at 9:57 am

Hi, you can check this link: http://www.khmath.com/magic-number-1089/

November 4th, 2015 at 4:52 am

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