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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


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#1 bonanova

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Posted 19 September 2007 - 08:47 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?

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The greatest challenge to any thinker is stating the problem in a way that will allow a solution.
- Bertrand Russell

#2 normdeplume

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Posted 19 September 2007 - 07:23 PM   Best Answer

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.
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#3 bonanova

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Posted 19 September 2007 - 08:48 PM

That's it.

Here's my method: [care to share yours?]
Spoiler for solution

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The greatest challenge to any thinker is stating the problem in a way that will allow a solution.
- Bertrand Russell

#4 normdeplume

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Posted 20 September 2007 - 08:54 AM

I am afraid, it was pretty much the 'brute force' method.

As you say 59 and 119 looked promising, both working upto and including 6, but 7 was the dificulty, so I tried each increment of 60 (179, 239, 299 ...), until I got to 419, which worked for 7 as well, (hooray) but not 9 or 10 (boo). Then using the same logic as before (59 being the first number that worked up the number 6, Increments of 60 always worked for the first 6 numbers. So 419 being the first that works for numbers up to 7 I incremented by 420.) So 419, 839, 1259, 1679, 2099, 2519.

Not a very elogant solution, but it passed a little time.

I assume, some mathemarical genius is going to post an easier way

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
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#5 ghyspran

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Posted 03 October 2007 - 02:25 AM

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.
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#6 brhan

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Posted 03 October 2007 - 03:54 PM

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?



Let X be the number. From the above properties of X, one can see that,
"X+1" is a multiple of the numbers from 2 to 10. As Bonanova is asking for the smallest number with such properties, "X+1" must be the LCM (least common multiple) of the numbers from 2 to 10 -- which is 2520.

And hence, X = 2520-1 = 2519.
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#7 brhan

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Posted 03 October 2007 - 03:57 PM

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?
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#8 brhan

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Posted 06 October 2007 - 05:50 PM

Not sure if people are checking the variation of the above question. I think it is better to post it as a new question.
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#9 vivek_t

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Posted 15 January 2008 - 06:51 AM

All we need to do is to take the LCM of all these numbers and subtract 1 from that..
LCM of 1 to 10 is 5x8x9x7=2520
-1 is 2519
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#10 PtGrill

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Posted 18 January 2008 - 10:15 PM

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?


A number that satisfies all of these properties is:
Spoiler for above problem

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