find the number

[Zapp]

- 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

[WhoDatMath]

Answer!

Spoiler! :

23815

[Zapp]

that's it. how did you solve it? I used a somehow complicated method, I want to see if there is an easier way

[WhoDatMath]

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.

[WhoDatMath]

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.

[xander]

I see what you did there.

xander

[Phoenix]

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.

[WhoDatMath]

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

[Zapp]

if there is, I think we would have noticed it while solving the puzzle

[Phoenix]

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

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

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.

[WhoDatMath]

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

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

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)

[Phoenix]

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

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

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. )

[WhoDatMath]

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