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 April 28th, 2013, 03:02 AM #1 Senior Member   Joined: Feb 2013 Posts: 114 Thanks: 0 Number system - proving 9 digit number not divisible by 5 Prove that there is no such 9 digit number in which every digit except zero occurs (once) and which ends in 5 can't be square ( digits are not to be repeated). Please suggest the solution of this question.
April 28th, 2013, 04:27 AM   #2
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Re: Number system - proving 9 digit number not divisible by

Quote:
 Originally Posted by sachinrajsharma Prove that there is no such 9 digit number in which every digit except zero occurs (once) and which ends in 5 can't be square ( digits are not to be repeated). Please suggest the solution of this question.
I don't see the connection between your subject line and the question you go on to ask.

 April 28th, 2013, 04:34 AM #3 Math Team     Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory Re: Number system - proving 9 digit number not divisible by Maybe he asks to prove that no 9 digit number follows these properties : 1) It is divisible by 5 2) Every integer from 1-9 are it's digits 3) The number is a perfect square I haven't ran a brute-force check yet but my heuristics says that if such number exist, then it must have the last two digits 2 & 5.
 April 28th, 2013, 04:59 AM #4 Math Team     Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory Re: Number system - proving 9 digit number not divisible by This is the possible list of such numbers I found using brute-force search : 100500625,102515625,104550625,106605625,108680625, 110775625,112890625,115025625, 117180625,119355625,121550625,123765625,126000625, 128255625,130530625,132825625, 135140625,137475625,139830625,142205625,144600625, 147015625,149450625,151905625, 154380625,156875625,159390625,161925625,164480625, 167055625,169650625,172265625, 174900625,177555625,180230625,182925625,185640625, 188375625,191130625,193905625, 196700625,199515625,202350625,205205625,208080625, 210975625,213890625,216825625, 219780625,222755625,225750625,228765625,231800625, 234855625,237930625,241025625, 244140625,247275625,250430625,253605625,256800625, 260015625,263250625,266505625, 269780625,273075625,276390625,279725625,283080625, 286455625,289850625,293265625, 296700625,300155625,303630625,307125625,310640625, 314175625,317730625,321305625, 324900625,328515625,332150625,335805625,339480625, 343175625,346890625,350625625, 354380625,358155625,361950625,365765625,369600625, 373455625,377330625,381225625, 385140625,389075625,393030625,397005625,401000625, 405015625,409050625,413105625, 417180625,421275625,425390625,429525625,433680625, 437855625,442050625,446265625, 450500625,454755625,459030625,463325625,467640625, 471975625,476330625,480705625, 485100625,489515625,493950625,498405625,502880625, 507375625,511890625,516425625, 520980625,525555625,530150625,534765625,539400625, 544055625,548730625,553425625, 558140625,562875625,567630625,572405625,577200625, 582015625,586850625,591705625, 596580625,601475625,606390625,611325625,616280625, 621255625,626250625,631265625, 636300625,641355625,646430625,651525625,656640625, 661775625,666930625,672105625, 677300625,682515625,687750625,693005625,698280625, 703575625,708890625,714225625, 719580625,724955625,730350625,735765625,741200625, 746655625,752130625,757625625, 763140625,768675625,774230625,779805625,785400625, 791015625,796650625,802305625, 807980625,813675625,819390625,825125625,830880625, 836655625,842450625,848265625, 854100625,859955625,865830625,871725625,877640625, 883575625,889530625,895505625, 901500625,907515625,913550625,919605625,925680625, 931775625,937890625,944025625, 950180625,956355625,962550625,968765625,975000625, 981255625,987530625,993825625, 1000140625,1006475625,1012830625,1019205625, Everyone of these has either two 5's or a 0. Hence, such number don't exist. I think a rigorous proofs follows from the fact that the last two digits are 2 and 5.
 April 28th, 2013, 12:29 PM #5 Math Team   Joined: Apr 2012 Posts: 1,579 Thanks: 22 Re: Number system - proving 9 digit number not divisible by Well, it SOUNDS like he wants wants only nine digit numbers where all of the digits 1,2,3,4,5,6,7,8,9 occur. No zeroes anywhere. Now, If all digits one through 9 occur at least once, then they all occur ONLY once and no special stipulation is required. Obviously, all (and only) those permutations of 1,2,3,4,6,7,8 and 9 with 5 tacked on the end WILL be divisible by 5. I can't process what "can't be square" means. If you are ruling out cases where the resultant nine digit number is a square, you will still end up with lots that meet the criteria, including the "first" case: 123467895, whose square root is greater than 111 and less than 112. So, I am still missing the drift OR the desired result is simply wrong. So I suppose Balarka's interpretation must be the correct one. What is desired is a proof that none of the numbers that ARE divisible by 5 are square. If you ARE a square and divisible by 5, then you must be divisible by 25. In THAT case, Balarka's brute force demo that all nine digit multiples of 25 either contain 0s or repeats, and specifically repeats of 5, which is kind of intriguing, does indeed prove the case.
 April 28th, 2013, 08:43 PM #6 Math Team     Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233 Re: Number system - proving 9 digit number not divisible by I agree so far with everything but would like to add, Every number that can be made using all digits 1 - 9 is divisible by 9 since the digital root of any such number is 9 Now, we are looking for a square that is also divisible by 5. A square divisible by 5 must be divisible by 25 so we are looking for a number of the form $9*25*k^2= (3*5*k)^2 = (15k)^2$ for some integer 740 < k < 2096 all even integers can be excluded at once since the 15 will make last digit zero Where to go next, i'm not sure. [color=#0000BF]mathbalarka[/color], I like your approach, numbers < 123456789 or numbers > 987654321 can be crossed off your list. That eliminates 16 candidates, a slight improvement but not much.
April 29th, 2013, 03:37 AM   #7
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Re: Number system - proving 9 digit number not divisible by

Quote:
 Originally Posted by agentredlum so we are looking for a number of the form $9*25*k^2= (3*5*k)^2 = (15k)^2$
This property is an interesting one, although it's not useful for brutal methods. I have to think a bit more to derive a rigorous proof I guess...

Quote:
 Originally Posted by agentredlum numbers < 123456789 or numbers > 987654321 can be crossed off your list. That eliminates 16 candidates
Yes, I didn't noticed this property, nice one agentredlum! This reduces the amount of checking needed to 200 candidates.

 April 29th, 2013, 05:49 AM #8 Math Team   Joined: Apr 2012 Posts: 1,579 Thanks: 22 Re: Number system - proving 9 digit number not divisible by Ok, building on agent's astute observation: Since whatever factors beyond 25 and 9 the number has must be odd squares, the number must ultimately be of the form: n = 900(x^2+x)+225 So, if you plug in any n of the relevant form into the following quadratic equation: x^2 + x + ((n-225)/900) = 0 Do you get any integral solutions? If not, then there is no such number. Still a bit brutish, however, ...

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# prove that a 9 digit number in which every digit except 0 zero occurs and which ends in 5, cannot be a square

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