66 replies to this topic

[woon]

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?

LCM? Lowest Common Multiplier?
Since the number is always remain (n-1) when devided by n ( n = 2 to 10), then it will be just the common multiplier of 2, 3, 4....to 9 and then -1.

LCM of (2, 3, 4,......,8,9)= 2520
2520 - 1 = 2519 !

then the next one will be 2(2520)-1 = 5039; 3(2520)-1 = 7559, etc....

[clao784]

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.

[ckreiter]

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?

[Romulus064]

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...

As has been said thousands of times on this post as if it was new:
Take the LCM of 1-12,by multiplying 2520 by 11 to get 27720
Minus 1 to add the remainder
end by getting:
27719

[cmgogo00]

2519

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

[Andrew1258]

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

[MRavi]

DEAR FRIENDS..

ANS IS 3628801

PLEASE CHECK AND REPLY..

RAVI

[mIk3y]

Spoiler for Answer

I found that 11 is the lowest answer.

[quote name='bonanova' date='19 September 2007 - 01:47 AM' timestamp='1190188021' post='4882']

[jerbil]

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

This problem may be solved as a simple variant of the Chinese Remainder Theorem, BN. If you do not have a book on Elementary Number Theory to hand, just Google it. The problem was first posed by the great physicist Dirac (in the 1930's I seem to remember, at a Mathematical symposium.)

[azjain]

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.