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#1 2016-05-12 19:24:08

chen.aavaz
Member
Registered: 2016-05-05
Posts: 34

Two-blocks number

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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#2 2016-05-13 01:29:29

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

Re: Two-blocks number

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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#3 2016-05-13 02:18:07

chen.aavaz
Member
Registered: 2016-05-05
Posts: 34

Re: Two-blocks number

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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#4 2016-05-13 03:48:33

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

Re: Two-blocks number

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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#5 2016-05-13 04:16:12

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

Re: Two-blocks number

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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#6 2016-05-13 05:07:27

chen.aavaz
Member
Registered: 2016-05-05
Posts: 34

Re: Two-blocks number

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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#7 2016-05-13 05:15:56

chen.aavaz
Member
Registered: 2016-05-05
Posts: 34

Re: Two-blocks number

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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#8 2016-05-13 05:23:54

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

Re: Two-blocks 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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#9 2016-05-13 07:32:14

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

Re: Two-blocks number

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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#10 2016-05-13 07:34:17

chen.aavaz
Member
Registered: 2016-05-05
Posts: 34

Re: Two-blocks number

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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#11 2016-05-13 07:46:39

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

Re: Two-blocks number

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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#12 2016-05-13 07:49:43

chen.aavaz
Member
Registered: 2016-05-05
Posts: 34

Re: Two-blocks number

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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#13 2016-05-13 07:51:30

chen.aavaz
Member
Registered: 2016-05-05
Posts: 34

Re: Two-blocks number

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

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#14 2016-05-13 07:59:54

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

Re: Two-blocks number

Oh, I see. I will keep looking smile

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#15 2016-05-13 08:28:04

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

Re: Two-blocks number

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

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#16 2016-05-13 08:34:07

chen.aavaz
Member
Registered: 2016-05-05
Posts: 34

Re: Two-blocks number

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 a is 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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#17 2016-05-13 15:28:28

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

Re: Two-blocks number

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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#18 2016-05-13 15:42:23

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

Re: Two-blocks number

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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#19 2016-05-13 15:49:20

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

Re: Two-blocks number

# 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.)


{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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#20 2016-05-13 19:54:33

chen.aavaz
Member
Registered: 2016-05-05
Posts: 34

Re: Two-blocks number

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:

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.

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#21 2016-05-13 23:46:08

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

Re: Two-blocks number

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.


{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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#22 2016-05-14 00:13:14

chen.aavaz
Member
Registered: 2016-05-05
Posts: 34

Re: Two-blocks number

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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#23 2016-05-14 02:22:42

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

Re: Two-blocks number

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.

Offline

#24 2016-05-14 02:40:04

chen.aavaz
Member
Registered: 2016-05-05
Posts: 34

Re: Two-blocks number

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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#25 2016-05-14 03:58:21

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

Re: Two-blocks number

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

,6 is to be appended to give a multiple of 49 . clearly it is 196.So we should look out for remainder 19. In my casio fx-825X calculator there is a button a b/c for integer division which gives dividend and remainder. Start with
= gives
remainder is not 19. To continue store it in memory and multiply by 10 ,look at remainder ,replace the number in memory by this number and go on multiplying by 10 until it gives decimal number.this is due to the ceiling on such operation.now recall the number in memory, write down integer part of division on paper (20408 in my calculation) and restart with remainder 8 and 0 to the right i.e 80 divide by 49 the same way from this batch I get integer as 1632 which you write on the right hand side of previous part(20408) and remainder 32 . continue the process until you get remainder of 19. Append 6 to the right and the dividend is 196/49=4 which is the last digit of number "a". i had done this much of work and posted it. I did not even calculate "b".I left it to chen.aavaz. He must have had horrible time for (a+b)^2.

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)


{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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