- 17 May '08 20:46 / 1 editWhen I sell my apples in the market I am very much into symmetry. I always put them in piles of the same amount in order to show them nicely.

So today I put them in 3 piles of the same number of apples in each pile. But I got one apple left.

I tried 5 piles, but the same happened, one apple too many to be able to divide them evenly.

I tried 7 piles, and 9 piles, with the same result, one apple over.

But when I divided them into 11 piles, I got it right. All of the 11 piles had the same number of apples in them. No apple too many.

How many apples did I bring to the market? - 17 May '08 21:21

I got the number 1,771,561. That's a lot of apples. So how long did it take you to sort all those apples? even if you sorted them at ~102.5 apples per second it would still take you all day*Originally posted by FabianFnas***When I sell my apples in the market I am very much into symmetry. I always put them in piles of the same amount in order to show them nicely.**

So today I put them in 3 piles of the same number of apples in each pile. But I got one apple left.

I tried 5 piles, but the same happened, one apple too many to be able to divide them evenly.

I tried 7 piles, ...[text shortened]... e same number of apples in them. No apple too many.

How many apples did I bring to the maket? - 17 May '08 21:34

How did you come up with this result?*Originally posted by strokem1***I got the number 1,771,561. That's a lot of apples. So how long did it take you to sort all those apples? even if you sorted them at ~102.5 apples per second it would still take you all day**

The correct answer gives no points if you don't show your calculation, as my teacher used to say... - 17 May '08 21:40 / 2 edits

Trial and error. I just began playing with powers of 11 checking to see if one was subtracted from the number if 5 would work first since it was the easiest to see, then 9/3, then if 7 would work... until I came up with 11^6 i believe it was...*Originally posted by FabianFnas***How did you come up with this result?**

The correct answer gives no points if you don't show your calculation, as my teacher used to say...

I was planning on actually working it out by hand but this method seemed to work so much better. I'll get the actual equations one of these days - 17 May '08 22:06

You didn't ask for minimum, but assuming that's what you meant ;*Originally posted by FabianFnas***When I sell my apples in the market I am very much into symmetry. I always put them in piles of the same amount in order to show them nicely.**

So today I put them in 3 piles of the same number of apples in each pile. But I got one apple left.

I tried 5 piles, but the same happened, one apple too many to be able to divide them evenly.

I tried 7 piles, ...[text shortened]... same number of apples in them. No apple too many.

How many apples did I bring to the market?

(3*5*7*9)+1=946 - 18 May '08 00:26 / 1 edit

1 Apple would be okay (gives remainder 1 when divided by 3,5,7,9), if 1 was divisible by 11. But 1 isn't divisible by 11, so we should add a muliple of lcm(3,5,7,9) [that is, the least common multiple of 3,5,7,9] to 1 to make it divisible by 11, since adding a multiple of lcm(3,5,7,9) to a number does not change it's remainder when divided by 3,5,7,9.*Originally posted by FabianFnas***When I sell my apples in the market I am very much into symmetry. I always put them in piles of the same amount in order to show them nicely.**

So today I put them in 3 piles of the same number of apples in each pile. But I got one apple left.

I tried 5 piles, but the same happened, one apple too many to be able to divide them evenly.

I tried 7 piles, ...[text shortened]... same number of apples in them. No apple too many.

How many apples did I bring to the market?

lcm(3,5,7,9)=lcm(5,7,9) [since 9 is divisible by 3] = 5*7*9 (since 5,7,9 are relatively prime in pairs) = 315.

To make calculations easier we compute modulo 11: so 315=7.

How many times do we add 7 to 1 to get a muliple of 11? 3 times: 3*7+1=22.

So the answer is 3*315+1=946. - 18 May '08 01:47

well if you put it that way... I guess I did over simplify my methods for finding an answer... I have an additional question, since we know that 946 and 1,771,561 work, how often do these "special numbers" occur?*Originally posted by David113***1 Apple would be okay (gives remainder 1 when divided by 3,5,7,9), if 1 was divisible by 11. But 1 isn't divisible by 11, so we should add a muliple of lcm(3,5,7,9) [that is, the least common multiple of 3,5,7,9] to 1 to make it divisible by 11, since adding a multiple of lcm(3,5,7,9) to a number does not change it's remainder when divided by 3,5,7,9.**

lc ...[text shortened]... s do we add 7 to 1 to get a muliple of 11? 3 times: 3*7+1=22.

So the answer is 3*315+1=946. - 18 May '08 02:00 / 1 edit

If you would like an infinite set of the special numbers, then I believe you can take, for example, 946^n (n=1,2,3,...). You could also take your number and raise it to powers, 1771561^n (n=1,2,3,...). Or I think more generally, any special number raised to powers n=1,2,3,....*Originally posted by strokem1***well if you put it that way... I guess I did over simplify my methods for finding an answer... I have an additional question, since we know that 946 and 1,771,561 work, how often do these "special numbers" occur?**

This works because every special number is of the form X+1, where X+1 is divisible by 11 and X is divisible by 3,5,7,9. But then (X+1)^n is also a special number because, first off, obviously it is also divisible by 11. Further, (X+1)^n always expands to something like X^n + C*X^(n-1) + C*X^(n-2) + ... + C*X + 1, which is of the form Y+1, where Y is obviously also divisible by 3,5,7,9 if X is (the 'C' just denotes that some non-negative integer constant is there).

Other than that, I'm not sure how "often" the special numbers occur. - 18 May '08 02:49

One apple too many?*Originally posted by FabianFnas***When I sell my apples in the market I am very much into symmetry. I always put them in piles of the same amount in order to show them nicely.**

So today I put them in 3 piles of the same number of apples in each pile. But I got one apple left.

I tried 5 piles, but the same happened, one apple too many to be able to divide them evenly.

I tried 7 piles, ...[text shortened]... same number of apples in them. No apple too many.

How many apples did I bring to the market?

GRANNY. - 18 May '08 03:06 / 3 editsmodular bash

let x be the number of apples

x=1 (3)

x=1 (5)

x=1 (7)

x=0 (11)

By the Chinese Remainder Theorem there is exactly one solution (mod 1155)

Again by the CRT x=1 (105)

We wish to find k such that 11|x => 11|(1+105k)

105k=10 (11)

6k=10 (11)

k=9 (11)

So x=1+945=946 (mod 1155)

Minimal solution is 946

EDIT: Whoops didnt see the 9 in the question. By the CRT there is one solution (mod 3465) cbf finding it.

EDIT2: I just realised, the solution 946 is 1 mod 9 anyway, so the solution is 946 (mod 3465)

EDIT3: Your question does not say put any restrictions, so there are an infinite number of solutions. You could have brought 946 apples, or brought 3465946 apples. Nobody knows. XP - 18 May '08 04:30
*Originally posted by FabianFnas*

So today I put them in 3 piles of the same number of apples in each pile. But I got one apple left.

I tried 5 piles, but the same happened, one apple too many to be able to divide them evenly.

I tried 7 piles, ...[text shortened]... same number of apples in them. No apple too many.

How many apples did I bring to the market?

3n+1=/=11

3n+1 = 22

5m+1=/=22

3n+1=/=33 or 44

3n+1 = 55

5m+1=/=55

Oh this is going to take forever. Forget it. - 18 May '08 05:01

1+ every (3+11n)th multiple of 315. For n=0;1;2;3.......*Originally posted by strokem1***well if you put it that way... I guess I did over simplify my methods for finding an answer... I have an additional question, since we know that 946 and 1,771,561 work, how often do these "special numbers" occur?**