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**chen.aavaz****Member**- Registered: 2016-05-05
- Posts: 37

Hi guys,

I need help with this puzzle please:

Find a natural number, which, if appended to the leftmost of another natural number, multiple of the first one, the new number that is formed equals to the square of the sum of the two initial numbers (the first one and its multiple).

The second number cannot be zero.

Just an attempt, but up to a certain point. Then I will leave it to you!

Suppose the numbers are a and b, with b=k*a, k>=1.

Let c be the new number which is formed by the other two.

c=(10^n)*a+b,

again n>=1 (n is the number of digits of a)

We therefore have: c=10^n*a+k*a = a*(10^n+k) = (a+b)^2

So: (a+k*a)^2=a*(10^n+k)-->[a(k+1)]^2=a*(10^n+k)-->a*(k^2+2*k+1)=10^n+k-->a*k^2+2*a*k+a-k=10^n

So a=(10^n+k)/(k^2+2*k+1) and a must be an integer. Is this possible?

*Last edited by chen.aavaz (2016-05-12 19:25:19)*

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**Relentless****Member**- Registered: 2015-12-15
- Posts: 624

Hi, I don't know if this was apparent already, and it's not a big clue, but I confirmed experimentally that at least one of *a*, *k* or *n* is greater than 100.

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**chen.aavaz****Member**- Registered: 2016-05-05
- Posts: 37

I see.

Maybe this is not the right approach to the problem, so I don't want to mislead you, but, if the final equation lies as in my previous message, we must determine the (possible) boundaries for k, n and therefore a, so as to start testing several values.

Just a few ideas:

Since the denominator is (k+1)^2, (k+1) must be a (double) factor of 10^n + k.

Can we deduce also a relationship between n, k and a, based on the fact that n = log[(c-b)/a]?

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

if appended to the leftmost of another natural number, multiple of the first one

Can you clear up what that means?

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**thickhead****Member**- Registered: 2016-04-16
- Posts: 1,086

Hi, chen.aavaz,

n should be the the number of digits of 'b" ,not "a".

I can only say that 'k" can not be an odd number.

So computational work is reduced by 50%

*Last edited by thickhead (2016-05-13 04:53:51)*

**{1}Vasudhaiva Kutumakam.{The whole Universe is a family.}(2)Yatra naaryasthu poojyanthe Ramanthe tatra Devataha{Gods rejoice at those places where ladies are respected.}**

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**chen.aavaz****Member**- Registered: 2016-05-05
- Posts: 37

It means that we have a natural number "a" and a second one, "b", which is a multiple of a (b=k*a) and we "stick" them together to form a third number "c", of which the right part is b and the left part is a. For example, 21 and 1638 (multiple of 21) and the new number is 211638. We are looking for a number with the property c = (a+b)^2. I know there exists.

bobbym wrote:

if appended to the leftmost of another natural number, multiple of the first one

Can you clear up what that means?

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**chen.aavaz****Member**- Registered: 2016-05-05
- Posts: 37

Yes, you are right about n.

Why k cannot be odd?

Also, from the properties of perfect squares, the denominator cannot end in 2,3,7 or 8.

But I don't want to mislead you because there might be an easier solution!

My understanding, though, is that the only way to solve it is to somehow limit the boundaries and do some tests.

I know that the number we are looking for is really big...

thickhead wrote:

Hi, chen.aavaz,

n should be the the number of digits of 'b" ,not "a".

I can only say that 'k" can not be an odd number.

So computational work is reduced by 50%

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**thickhead****Member**- Registered: 2016-04-16
- Posts: 1,086

**{1}Vasudhaiva Kutumakam.{The whole Universe is a family.}(2)Yatra naaryasthu poojyanthe Ramanthe tatra Devataha{Gods rejoice at those places where ladies are respected.}**

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**Relentless****Member**- Registered: 2015-12-15
- Posts: 624

Hi! I found three answers so far:

1. Set k=8, n=8, a=1234568, b=9876544,

c=123456809876544

2. Set k=10, n=12, a=8264462810, b=82644628100,

c=8264462810082644628100

3. Set k=40, n=10, a=5948840, b=237953600,

c=59488400237953600

Further answer:

4. Set k=188, n=6, a=28, b=5264,

c=28005264

*Last edited by Relentless (2016-05-13 07:56:35)*

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**chen.aavaz****Member**- Registered: 2016-05-05
- Posts: 37

Why do you put the zero in-between? The two numbers a and b must be appended without any other digit in the middle.

*Last edited by chen.aavaz (2016-05-13 07:38:26)*

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**Relentless****Member**- Registered: 2015-12-15
- Posts: 624

Hi, I don't understand... is the problem not just to choose three positive integers such that a=(k+10^n)/(k+1)^2 ?

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**chen.aavaz****Member**- Registered: 2016-05-05
- Posts: 37

The problem is as stated in my first post. Also read my reply to Bobbym.

Relentless wrote:

Hi, I don't understand... is the problem not just to choose three positive integers such that a=(k+10^n)/(k+1)^2 ?

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**chen.aavaz****Member**- Registered: 2016-05-05
- Posts: 37

@Relentless: what you have done would be correct if there weren't the 0 in the middle!

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**Relentless****Member**- Registered: 2015-12-15
- Posts: 624

Oh, I see. I will keep looking

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**Relentless****Member**- Registered: 2015-12-15
- Posts: 624

Having trouble finding any more integer results. It seems likely that *a* is bigger than 10^12....

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**chen.aavaz****Member**- Registered: 2016-05-05
- Posts: 37

Very likely.

We must somehow set the boundaries.

There is also one more condition: n is the number of digits of a*k, thus a, n and k are not totally independent.

Relentless wrote:

Having trouble finding any more integer results. It seems likely that

ais bigger than 10^12....

Let me know if the below are correct:

Let's examine k=1:

It must be a*10^n+a=(2a)^2 --> a*(10^n+1)=4*a^2 --> 10^n+1 must be divisible by 4, which is not possible, therefore k cannot be 1.

Similarly it cannot be 2 because then 10^n+2 must be divisible by 9, impossible.

It cannot be 3 because 10^n+3 must be divisible by 16,

neither 4, neither 5.

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

I do not think there are any solutions.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**thickhead****Member**- Registered: 2016-04-16
- Posts: 1,086

chen.aavaz wrote:

Yes, you are right about n.

Why k cannot be odd?Also, from the properties of perfect squares, the denominator cannot end in 2,3,7 or 8.

But I don't want to mislead you because there might be an easier solution!

My understanding, though, is that the only way to solve it is to somehow limit the boundaries and do some tests.

I know that the number we are looking for is really big...thickhead wrote:Hi, chen.aavaz,

n should be the the number of digits of 'b" ,not "a".

I can only say that 'k" can not be an odd number.

So computational work is reduced by 50%

If k is odd, 1+k & (1+k)^2 are even but 10^n+k is odd. An odd number can not be an integrral multiple of an even number.

**{1}Vasudhaiva Kutumakam.{The whole Universe is a family.}(2)Yatra naaryasthu poojyanthe Ramanthe tatra Devataha{Gods rejoice at those places where ladies are respected.}**

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**thickhead****Member**- Registered: 2016-04-16
- Posts: 1,086

# relentless,

I had also found this number among others but the number of digits in 'b" did not match "n", i fact fell short by 1 in several cases. only thereafter I made analysis about the number of digits.

k=8, n=8, a=1234568, b=9876544 ( of 7 digits only.)

(2)Yatra naaryasthu poojyanthe Ramanthe tatra Devataha

{Gods rejoice at those places where ladies are respected.}

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**chen.aavaz****Member**- Registered: 2016-05-05
- Posts: 37

Got it, thanks.

I also think k must have more than 2 digits.

thickhead wrote:

chen.aavaz wrote:Yes, you are right about n.

Why k cannot be odd?Also, from the properties of perfect squares, the denominator cannot end in 2,3,7 or 8.

But I don't want to mislead you because there might be an easier solution!

My understanding, though, is that the only way to solve it is to somehow limit the boundaries and do some tests.

I know that the number we are looking for is really big...thickhead wrote:

n should be the the number of digits of 'b" ,not "a".

I can only say that 'k" can not be an odd number.

So computational work is reduced by 50%If k is odd, 1+k & (1+k)^2 are even but 10^n+k is odd. An odd number can not be an integrral multiple of an even number.

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**thickhead****Member**- Registered: 2016-04-16
- Posts: 1,086

hurrah!! Found a number.

a=20408163265306122448979591836734694

k=6

utilities used;scientific calculator, pen ,paper and some grey cells.

Please check whether it satisfies all the conditions.

(2)Yatra naaryasthu poojyanthe Ramanthe tatra Devataha

{Gods rejoice at those places where ladies are respected.}

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**chen.aavaz****Member**- Registered: 2016-05-05
- Posts: 37

YESSSSSSSSSSSSSSSSSS!!!

thickhead wrote:

hurrah!! Found a number.

a=20408163265306122448979591836734694

k=6

utilities used;scientific calculator, pen ,paper and some grey cells.

Please check whether it satisfies all the conditions.

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

Hi;

Can not verify that. I do not think that is correct. What is your a and b?

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**chen.aavaz****Member**- Registered: 2016-05-05
- Posts: 37

It is.

a=20408163265306122448979591836734694

k=6

b=122448979591836734693877551020408164

c=20408163265306122448979591836734694122448979591836734693877551020408164=

(142857142857142857142857142857142858)^2=(20408163265306122448979591836734694+122448979591836734693877551020408164)^2

bobbym wrote:

Hi;

Can not verify that. I do not think that is correct. What is your a and b?

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**thickhead****Member**- Registered: 2016-04-16
- Posts: 1,086

Hi chen.aavaz,

I am glad that it fits well.

I found it by actual division of (10^n+k)/(k+1)^2

since k has to be low to meet demands of n I chose least possible value of k=6. denom=49

We do not know n.we know that with last remainder of

If you find calculator too tiresome you can try excel.

start with a series 100. 1000.10000,100000 on a column "a"say. in the next column use=quotient(a1,49) and in column c =mod(a1,49). Copy the formula to lower cells. however after a few cells down the calculation fails. Note down the last quotient and erase all failed values. and start with new series say 80,800,8000 etc (if 8 was the last remainder) continue the process until you spot remainder=19.

*Last edited by thickhead (2016-05-16 04:43:08)*

(2)Yatra naaryasthu poojyanthe Ramanthe tatra Devataha

{Gods rejoice at those places where ladies are respected.}

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