How many apples?

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, 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?

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?

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 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

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...

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...

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...

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

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?

You didn't ask for minimum, but assuming that's what you meant ;
(3*5*7*9)+1=946

Luskin, numbers divisible by 9 are inherently divisible by 3.
So your solution for the lowest can be changed to
(5*7*9)+1=316.

I'm trying to see if I can get less.

Originally posted by tamuzi
Luskin, numbers divisible by 9 are inherently divisible by 3.
So your solution for the lowest can be changed to
(5*7*9)+1=316.

I'm trying to see if I can get less.

But 316 is not divisible by 11.

knew I was missing something

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?

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.

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.

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.

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 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?

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,....

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.

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?

One apple too many?

GRANNY.

modular 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

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?

x = 3n+1 = 5m+1 = 7p+1 = 9q+1 = 11r

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.

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?

1+ every (3+11n)th multiple of 315. For n=0;1;2;3.......