Help! A remainder is chasing me

68 posts in this topic

Posted · Report post

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?

This time X-1 should be the LCM and hence the number is 2521

0

Share this post


Link to post
Share on other sites

Posted · Report post

This time X-1 should be the LCM and hence the number is 2521

Nope ... 2521 is not divisible by 11.

0

Share this post


Link to post
Share on other sites

Posted (edited) · Report post

Nope ... 2521 is not divisible by 11.

New solution

The number is 25201.

My solution is that the number should be X=2520Y+1 and for some integer Y X should be divisible by 11. Is there some simpler method?

Edited by imran
0

Share this post


Link to post
Share on other sites

Posted · Report post

New solution

The number is 25201.

My solution is that the number should be X=2520Y+1 and for some integer Y X should be divisible by 11. Is there some simpler method?

that is correct. good job ... I just managed to do it the brute forth method.

0

Share this post


Link to post
Share on other sites

Posted · Report post

How can the answer be......

2519

0

Share this post


Link to post
Share on other sites

Posted · Report post

Here is my answer to the original puzzle.

My answer is -1. No matter what integer you divide it by (except of course zero) the remainder is one less than the divisor.

0

Share this post


Link to post
Share on other sites

Posted (edited) · Report post

#include <cstdlib>

#include <iostream>

using namespace std;

bool mod(unsigned int);

int main() {

	unsigned int i;

	for( i = 0; !mod(i); ++i);cout << i << endl;

	system("PAUSE");

	return EXIT_SUCCESS;

}


bool mod(unsigned int i) {

	for( unsigned int j = 2; j <= 10; ++j)

		if( i % j != (j - 1) )

			return false;

	return true;

}

Or, as my dad put it, (5*7*8*9)-1=2519

Damn engineers.

Edited by OstermanA
0

Share this post


Link to post
Share on other sites

Posted · Report post

If we're looking for n where n divided by x (x being each number 2-10) leaves a remainder of x-1, n+1 divided by each number 2-10 leaves no remainder. To find n+1, we must simply multiply all prime factors of 2-10, excluding any duplicates. This is equal to 2*2*2*3*3*5*7 = 2520, then just subtract one to get the answer of 2519.

thats right.

we are looking for (lcm of 1 to 10 )-1=2520-1=2519

0

Share this post


Link to post
Share on other sites

Posted · Report post

Here is my answer to the original puzzle.
My answer is -1. No matter what integer you divide it by (except of course zero) the remainder is one less than the divisor.

Since 2519 + 2520n where n is any integer, has the correct remainders, your answer fits at least part of the OP.

But you overlook the part of not being able to find a smaller answer.

If negative numbers are included, a smaller answer can always be found: for example, -2521 is less than your answer.

Therefore, negative numbers do not answer the OP.

0

Share this post


Link to post
Share on other sites

Posted · Report post

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?

Let X be the answer to the puzzle where X = Y + 1, Y is a multiple of 2 to 10

=> The LCM of 2-10 is 2520.

=> X = 2520Y + 1.

However, X mod 11=0

=> 2520 mod 11=1

=> Thus (2520 * 10) mod 11 = 10

=> Thus [(2520 * 10) + 1] mod 11 = 10 + 1 = 11 = 0

Therefore X = 25201

0

Share this post


Link to post
Share on other sites

Posted · Report post

There is another answer, which is -1. For example -1/9 = (-9+8)/9= -1 with remainder 8

0

Share this post


Link to post
Share on other sites

Posted · Report post

In the spirit of ATL, I present to you my method, in C#, for calculating the answer by brute force, without taking into account LCM's. I'm curious how much smaller this could be done.

int GetNumber() {

int i=1, n=1; bool r=false;

for (;!r;){if(i%n==(n-1)){n=n%11+1;r=n==11;}else{i++;n=1;}}return i;}

Using a while statement instead might save a few spaces, but the biggest space saver I would be changing n=n%11+1 to n++. Both increment n by 1 but n++ is much more logical. I don't really understand why you wouldn't use it. I'm sure there are more ways to reduce the code as well.

0

Share this post


Link to post
Share on other sites

Posted · Report post

my answer is 29

check it out :P

0

Share this post


Link to post
Share on other sites

Posted · Report post

A number that satisfies all of these properties is:

25201

I did not get it. 2519 is divisible by 11

0

Share this post


Link to post
Share on other sites

Posted (edited) · Report post

Since negative numbers are not excluded by the problem as posed, the answer is negative infinity - since any other solution has another number smaller than it. This doesn't fit well with the statement 'it is not a small number', but that is a subjective description.

2519

-1

-2521

-5041

-7561

-10081

-12601

-15121

-17641

etc.

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?

Edited by xucam
0

Share this post


Link to post
Share on other sites

Posted · Report post

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

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....

0

Share this post


Link to post
Share on other sites

Posted (edited) · Report post

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
0

Share this post


Link to post
Share on other sites

Posted · Report post

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?

0

Share this post


Link to post
Share on other sites

Posted · Report post

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

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

0

Share this post


Link to post
Share on other sites

Posted (edited) · Report post

2519

Edited by cmgogo00
0

Share this post


Link to post
Share on other sites

Posted · Report post

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

0

Share this post


Link to post
Share on other sites

Posted · Report post

DEAR FRIENDS..

ANS IS 3628801

PLEASE CHECK AND REPLY..

RAVI

0

Share this post


Link to post
Share on other sites

Posted · Report post

I found that 11 is the lowest answer.

0

Share this post


Link to post
Share on other sites

Posted · Report post

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>

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.)

0

Share this post


Link to post
Share on other sites

Posted (edited) · Report post

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
0

Share this post


Link to post
Share on other sites

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!


Register a new account

Sign in

Already have an account? Sign in here.


Sign In Now

  • Recently Browsing   0 members

    No registered users viewing this page.