## find the number

Have an interesting puzzle? Let's hear it!

### find the number

- it is a five digit number
- the sum of the first and second digit is the fifth
- the first OR the second digit is twice as large as the fourth
- the third digit is 8
- if you read the number from right to left, it is 28017 bigger

just to avoid confusion, the digit on the left is the first
Q.E.D. , or not?
Zapp
University

Posts: 124
Joined: Thu Feb 10, 2011 3:49 pm

### Re: find the number

Spoiler! :
23815
WhoDatMath
Mathlete

Posts: 78
Joined: Thu Feb 10, 2011 4:53 pm

### Re: find the number

that's it. how did you solve it? I used a somehow complicated method, I want to see if there is an easier way
Q.E.D. , or not?
Zapp
University

Posts: 124
Joined: Thu Feb 10, 2011 3:49 pm

### Re: find the number

I'm sure I didn't do it simpler than you did. Called the number ABCDE.

Made the assumption that A=2D. If B=2D instead I should find a contradiction and I'll start over.
C=8.
A+B=E

Set up EDCBA-ABCDE=28017. Since A is 2D, D<5. D-B is 8 meaning there has to be a carried digit. So D+10-B=8, or B=D+2.
Since B=D+2, the 1 in 28017 doesn't make sense, should be 2. So there's a carried digit, meaning that A+10-E=7 --> E=A+3 --> E=2D+3.

Substitution gives A+B=E --> 2D+D+2=2D+3 --> D=1.
A=2D=2.
E=2D+3=5.
B=E-A=5-2=3.
C=8.

23815.
Last edited by WhoDatMath on Fri Feb 11, 2011 2:27 pm, edited 1 time in total.
WhoDatMath
Mathlete

Posts: 78
Joined: Thu Feb 10, 2011 4:53 pm

### Re: find the number

Really quick, here's how you know that it's not B that equals twice D.

Subtraction gets us D-B=8, with D still <5, so D+10-B=8 --> B=D+2.

Substitution: B=2D --> 2D=D+2 --> D=2.

In the subtraction,

B-D=1
B=D+1
2D=D+1
4=3
QED.

We'll call this WhoDat's Lemma.
WhoDatMath
Mathlete

Posts: 78
Joined: Thu Feb 10, 2011 4:53 pm

### Re: find the number

I see what you did there.

xander

xander
University

Posts: 154
Joined: Fri Feb 11, 2011 12:14 am
Location: Sparks, NV, USA

### Re: find the number

WhoDatMath wrote:I'm sure I didn't do it simpler than you did. Called the number ABCDE.

Made the assumption that A=2D. If B=2D instead I should find a contradiction and I'll start over.
C=8.
A+B=E

Set up EDCBA-ABCDE=28017. Since A is 2D, D<5. D-B is 8 meaning there has to be a carried digit. So D+10-B=8, or B=D+2.
Since B=D+2, the 1 in 28017 doesn't make sense, should be 2. So there's a carried digit, meaning that A+10-E=7 --> E=A+3 --> E=2D+3.

Substitution gives A+B=E --> 2D+D+2=2D+3 --> D=1.
A=2D=2.
E=2D+3=5.
B=E-A=5-2=3.
C=8.

23815.

Nota bene: if you were asked to find all numbers which satisfy given rules, your solution wouldn't be complete. However, you were asked to find any number, so your solution works.
Phoenix
Elementary School

Posts: 13
Joined: Sat Feb 12, 2011 6:14 pm

### Re: find the number

Is there another number that satisfies the rules? I don't see how.
WhoDatMath
Mathlete

Posts: 78
Joined: Thu Feb 10, 2011 4:53 pm

### Re: find the number

if there is, I think we would have noticed it while solving the puzzle
Q.E.D. , or not?
Zapp
University

Posts: 124
Joined: Thu Feb 10, 2011 3:49 pm

### Re: find the number

WhoDatMath wrote:Is there another number that satisfies the rules? I don't see how.

There isn't, but you didn't prove that. You said:

WhoDatMath wrote:Made the assumption that A=2D. If B=2D instead I should find a contradiction and I'll start over.

And you found one answer. But what if there is a number that satisfies the case in which B=2D? You didn't complete that case.
Phoenix
Elementary School

Posts: 13
Joined: Sat Feb 12, 2011 6:14 pm

### Re: find the number

Phoenix wrote:
WhoDatMath wrote:Is there another number that satisfies the rules? I don't see how.

There isn't, but you didn't prove that. You said:

WhoDatMath wrote:Made the assumption that A=2D. If B=2D instead I should find a contradiction and I'll start over.

And you found one answer. But what if there is a number that satisfies the case in which B=2D? You didn't complete that case.

Really? Check my next post, man. I proved in a separate post that no number can satisfy B=2D. (WhoDat's Lemma)
WhoDatMath
Mathlete

Posts: 78
Joined: Thu Feb 10, 2011 4:53 pm

### Re: find the number

WhoDatMath wrote:
Phoenix wrote:
WhoDatMath wrote:Is there another number that satisfies the rules? I don't see how.

There isn't, but you didn't prove that. You said:

WhoDatMath wrote:Made the assumption that A=2D. If B=2D instead I should find a contradiction and I'll start over.

And you found one answer. But what if there is a number that satisfies the case in which B=2D? You didn't complete that case.

Really? Check my next post, man. I proved in a separate post that no number can satisfy B=2D. (WhoDat's Lemma)

True, that covers it... Somehow, that post wasn't there when I was reading the topic.

(Why is it so hard for me to follow this forum? This is not my first time to look over something. )
Phoenix
Elementary School

Posts: 13
Joined: Sat Feb 12, 2011 6:14 pm

### Re: find the number

It's all good man. I added the second post after thinking "You know, somebody's gonna come along and say there could be another number..." So thanks for proving that the second post was worth the time and effort
WhoDatMath
Mathlete

Posts: 78
Joined: Thu Feb 10, 2011 4:53 pm