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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  
Math Team Joined: Apr 2012 Posts: 1,579 Thanks: 22  Re: Number system  proving 9 digit number not divisible by Quote:
 
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 19 are it's digits 3) The number is a perfect square I haven't ran a bruteforce 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 bruteforce 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: 3,330 Thanks: 219  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 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  
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 Quote:
Quote:
 
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 + ((n225)/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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