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# Help! A remainder is chasing me

Best Answer normdeplume , 19 September 2007 - 07:23 PM

My answer is below, is this the answer you were looking for or is there a lower one?

Spoiler for solution

Just because your voice reaches halfway around the world doesn't mean you are wiser than when it reached only to the end of the bar. Go to the full post

67 replies to this topic

### #41 woon

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Posted 12 November 2008 - 10:41 AM

I just found a number with an interesting property:

When I divide it by 2, the remainder is 1.
When I divide it by 3, the remainder is 2.
When I divide it by 4, the remainder is 3.
When I divide it by 5, the remainder is 4.
When I divide it by 6, the remainder is 5.
When I divide it by 7, the remainder is 6.
When I divide it by 8, the remainder is 7.
When I divide it by 9, the remainder is 8.
When I divide it by 10, the remainder is 9.

It's not a small number, but it's not really big, either.
When I looked for a smaller number with this property I couldn't find one.

Can you find it?

The method which I think most of the students can understand is to use the
Spoiler for Did you still remember?

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### #42 clao784

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Posted 03 February 2009 - 06:25 PM

Hi. numbers that fit that condition are:

2519, 5039, 7559, 10079, 12599, 15119, 17639, 20159, 22679, 25199, 27719, 30239, 32759, 35279, 37799, 40319, 42839, 45359, 47879, 50399, 52919, 55439, 57959, 60479, 62999, 65519, 68039, 70559, 73079, 75599, 78119, 80639, 83159, 85679, 88199, 90719, 93239, 95759, 98279, 100799, 103319, 105839, 108359, 110879, 113399, 115919, 118439, 120959, 123479, 125999, 128519, 131039, 133559, 136079, 138599, 141119, 143639, 146159, 148679, 151199, 153719, 156239, 158759, 161279, 163799, 166319, 168839, 171359, 173879, 176399, 178919, 181439, 183959, 186479, 188999, 191519, 194039, 196559, 199079, 201599, 204119, 206639, 209159, 211679, 214199, 216719, 219239, 221759, 224279, 226799, 229319, 231839, 234359, 236879, 239399, 241919, 244439, 246959, 249479, 251999, 254519, 257039, 259559, 262079, 264599, 267119, 269639, 272159, 274679, 277199, 279719, 282239, 284759, 287279, 289799, 292319, 294839, 297359, 299879, 302399, 304919, 307439, 309959, 312479, 314999, 317519, 320039, 322559, 325079, 327599, 330119, 332639, 335159, 337679, 340199, 342719, 345239, 347759, 350279, 352799, 355319, 357839, 360359, 362879, 365399, 367919, 370439, 372959, 375479, 377999, 380519, 383039, 385559, 388079, 390599, 393119, 395639, 398159, 400679, 403199, 405719, 408239, 410759, 413279, 415799, 418319, 420839, 423359, 425879, 428399, 430919, 433439, 435959, 438479, 440999, 443519, 446039, 448559, 451079, 453599, 456119, 458639, 461159, 463679, 466199, 468719, 471239, 473759, 476279, 478799, 481319, 483839, 486359, 488879, 491399, 493919, 496439, 498959, 501479, 503999, 506519, 509039, 511559, 514079, 516599, 519119, 521639, 524159, 526679, 529199, 531719, 534239, 536759, 539279, 541799, 544319, 546839, 549359, 551879, 554399, 556919, 559439, 561959, 564479, 566999, 569519, 572039, 574559, 577079, 579599, 582119, 584639, 587159, 589679...

you can check it, you can't find a number minor than 2519, you add 2520 each time and each number will fit the same condition!

Edited by clao784, 03 February 2009 - 06:27 PM.

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### #43 ckreiter

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Posted 26 February 2009 - 04:48 PM

I wrote an excel spreadsheet that calculated it for me. The number you are referring to is 2519 but here are a few larger numbers that work as well....
5039, 7559, 10079, 12599, 15119, 17639, 20159

I just found a number with an interesting property:

When I divide it by 2, the remainder is 1.
When I divide it by 3, the remainder is 2.
When I divide it by 4, the remainder is 3.
When I divide it by 5, the remainder is 4.
When I divide it by 6, the remainder is 5.
When I divide it by 7, the remainder is 6.
When I divide it by 8, the remainder is 7.
When I divide it by 9, the remainder is 8.
When I divide it by 10, the remainder is 9.

It's not a small number, but it's not really big, either.
When I looked for a smaller number with this property I couldn't find one.

Can you find it?

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### #44 Romulus064

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Posted 27 February 2009 - 01:17 AM

I think you should include dividing 11 to get a remainder of 10 and dividing 12 to get a remainder of 11

Spoiler for Well if you really want to take it farther...

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### #45 cmgogo00

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Posted 09 September 2009 - 07:38 PM

2519

Edited by cmgogo00, 09 September 2009 - 07:43 PM.

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### #46 Andrew1258

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Posted 10 September 2009 - 12:47 AM

Another variation of the above problem is as follows:

I just found a number with an interesting property:

When I divide it by 2, the remainder is 1.
When I divide it by 3, the remainder is 1.
When I divide it by 4, the remainder is 1.
When I divide it by 5, the remainder is 1.
When I divide it by 6, the remainder is 1.
When I divide it by 7, the remainder is 1.
When I divide it by 8, the remainder is 1.
When I divide it by 9, the remainder is 1.
When I divide it by 10, the remainder is 1.

But when I divide it by 11, the remainder is 0 (just to make the procedure a little bit different).

Can you find it?

Any number of the form

2X3X4X5X6X7X8X9XN -1

where N is any integer will satisfy the first 9 conditions, but since any number divided by 8 giving remainder 1 will also give the remainder 1 when divided by 2 or 4, and any number divided by 9, giving remainder 1 will also yield remainder 1 when divided by 3 therefore the number could be reduced to :

8X9X5X7n - 1 = 2520n -1

to be divisible by 11 we rerite it as:

2520n - 1 = 2530n - (10n+1)

2530n being a multiple of 11, we only have to fine n to make 10n +1 a multiple of 11

it's easy to see

n = 1 + 11m
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### #47 MRavi

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Posted 22 October 2009 - 03:40 PM

DEAR FRIENDS..

ANS IS 3628801

PLEASE CHECK AND REPLY..

RAVI
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### #48 mIk3y

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Posted 24 October 2009 - 01:40 AM

Spoiler for Answer
[quote name='bonanova' date='19 September 2007 - 01:47 AM' timestamp='1190188021' post='4882']
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### #49 jerbil

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Posted 24 October 2009 - 09:16 AM

That's it.

Here's my method: [care to share yours?]
<div style="margin:20px; margin-top:5px">
<div class="smallfont" style="margin-bottom:2px">Spoiler for solution: <input type="button" value="Show" style="width:45px;font-size:10px;margin:0px;padding:0px;" onClick="if (this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display != '') { this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display = ''; this.innerText = ''; this.value = 'Hide'; } else { this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display = 'none'; this.innerText = ''; this.value = 'Show'; }">
</div><div class="alt2" style="margin: 0px; padding: 6px; border: 1px inset;"><div style="display: none;">The number has to end in 9.
Looked brute force for small numbers.
59 and 119 were promising, but no cigar.

Then looked for agreement among
39 + multiples of 40,
69 + multiples of 70 and
89 + multiples of 90
Smallest one was 2519.

Still think of this as kind of brute force.
Maybe there is no elegant solution.</div></div></div>

Spoiler for Elegant Solution

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### #50 azjain

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Posted 03 January 2010 - 12:02 PM

Let the number is X.
Now consider the number X+1.
As the remainder for X/2 is 1, it follows X+1 is completely divisible by 2.
As the remainder for X/3 is 2, it follows X+1 is completely divisible by 3. and so on...

So X+1 is completely divisible by 2,3...10.
and the lowest such number has to be LCM of 2,3,...10 = 2520.

So X = 2520 - 1 = 2519.

Other number would be 2520n - 1, where n = 1,2,3...

Edited by azjain, 03 January 2010 - 12:05 PM.

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