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The sum of the digits of a two digit number is 11. When 28 is subtracted from the number, the digits become equal. Find the number.

Here is what I am doing:

I want to know if there is a way that does not involve trying all values of x one by one.

Is there a second equation I can form?

'And fun? If maths is fun, then getting a tooth extraction is fun. A viral infection is fun. Rabies shots are fun.'

'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'

I'm not crazy, my mother had me tested.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 105,268

Hi;

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.****No great discovery was ever made without a bold guess. **

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Any other method? That can be done using pen and paper?

'And fun? If maths is fun, then getting a tooth extraction is fun. A viral infection is fun. Rabies shots are fun.'

'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'

I'm not crazy, my mother had me tested.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 105,268

Use your head here a bit...

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.****No great discovery was ever made without a bold guess. **

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Yes, I tried banging my head on the desk. It hurts.

'And fun? If maths is fun, then getting a tooth extraction is fun. A viral infection is fun. Rabies shots are fun.'

'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'

I'm not crazy, my mother had me tested.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 105,268

Well then stop that immediately or use someone else's head.

Why do you think the problem only is a 2 digit number?

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.****No great discovery was ever made without a bold guess. **

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Please rephrase that question. I cannot purse it.

'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'

I'm not crazy, my mother had me tested.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 105,268

It is a two digit number so that you will not have to try so many!

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.****No great discovery was ever made without a bold guess. **

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Are you saying that there is an Intelligent Design or Creationism in the question?

'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'

I'm not crazy, my mother had me tested.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 105,268

Not an intelligent design but some design nonetheless.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.****No great discovery was ever made without a bold guess. **

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That is like hacking the mind of the examiner. Is it allowed?

'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'

I'm not crazy, my mother had me tested.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 105,268

A short while back I was discussing that with Agnishom version 1.0. He did not seem clever enough to understand that problem solving is war and any and all means are fair. This mostly applies to school and book problems. In short, yes hacking the mind is a powerful tool. Perhaps, Agnishom version 2.0 is smarter.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.****No great discovery was ever made without a bold guess. **

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**zetafunc.****Guest**

Agnishom wrote:

Any other method? That can be done using pen and paper?

You had the right idea.

Consider your equation, 9y - 17 = 11x. What can you say about that equation, mod 11?

**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 7,651

I did this. Still involves some trial I'm afraid.

Let x = 10a + b and x - 28 = 11c where a, b and c are whole numbers under 10.

b = 11 - a therefore x = 10a + 11 -a = 9a + 11

So

28 + 11c = 9a + 11

Therefore 9a - 28 must be a multiple of 11.

a must be > 3

So I tried a = 4, 5, 6, 7, 8 got it. Why didn't I start with 9 and work down.

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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zetafunc. wrote:

Agnishom wrote:Any other method? That can be done using pen and paper?

You had the right idea.

Consider your equation, 9y - 17 = 11x. What can you say about that equation, mod 11?

Please tell me more. I do not know much about modular equations

'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'

I'm not crazy, my mother had me tested.

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Bob, Isn't that the same solution as mine?

'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'

I'm not crazy, my mother had me tested.

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**zetafunc.****Guest**

bob bundy wrote:

I did this. Still involves some trial I'm afraid.

Let x = 10a + b and x - 28 = 11c where a, b and c are whole numbers under 10.

b = 11 - a therefore x = 10a + 11 -a = 9a + 11

So

28 + 11c = 9a + 11

Therefore 9a - 28 must be a multiple of 11.

a must be > 3

So I tried a = 4, 5, 6, 7, 8 got it. Why didn't I start with 9 and work down.

Bob

I don't think any trial and error is needed. Although it's probably quicker just to plug in a few numbers, you can use Agnishom's equation;

9x - 17 = 11z, for some positive integers x, z. Thus, working in modulo 11;

By the Euclidean algorithm,

.Hence:

Hence, x = 8, y = 3 is a solution.

**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 7,651

Well it was hidden so I wouldn't know.

Just had a look. It's very similar but I had less trials. I think your method is fine. Don't think you can get straight to a single solution.

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**zetafunc.****Guest**

I used different variables to Agnishom's post -- apologies. The principle is the same however.

**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 7,651

Neat solution. OK I stand corrected.

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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Hi zetafunc.,

What does the bar over the digits mean?

'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'

I'm not crazy, my mother had me tested.

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**ShivamS****Member**- Registered: 2011-02-07
- Posts: 3,648

Repeating decimal, sample mean, negation in boolean algebra, sometimes a vector (generally it's an arrow) or a complex conjugate.

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zetafunc. wrote:

By the Euclidean algorithm,

.

Please tell me that algorithm.

'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'

I'm not crazy, my mother had me tested.

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**zetafunc.****Guest**

Agnishom wrote:

Hi zetafunc.,

What does the bar over the digits mean?

ShivamS wrote:

Repeating decimal, sample mean, negation in boolean algebra, sometimes a vector (generally it's an arrow) or a complex conjugate.

Not in this context. Here,

denotes the set of elements congruent to a. More precisely:Let n be a fixed positive integer. For a,b ∈ **Z**, we write a ≡ b (mod n) if b - a is a multiple of n and say a is *congruent* to b.

**ShivamS****Member**- Registered: 2011-02-07
- Posts: 3,648

The Euclidean algorithm is a way to find the greatest common divisor of two positive integers.

Page 11 here: http://editorialdinosaurio.files.wordpress.com/2012/03/itn-niven.pdf

I couldn't explain it better my self.

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