Please help me solve and explain what 1886 equals in this:

7746=1

4509=2

6003=1

3669=3

7681=3

8898=7

9243=1

8285=4

5439=1

2282=2

1886=?

Results 1 to 7 of 7

- Mar 25th 2014, 03:57 PM #1

- Joined
- Mar 2014
- From
- Oregon
- Posts
- 3

- Mar 25th 2014, 05:49 PM #2

- Joined
- Nov 2010
- Posts
- 2,971
- Thanks
- 1141

## Re: Need help please

Those numbers don't equal the values you claim. Do you mean

$f(7746) = 1$

$f(4509) = 2$

...

...

...

$f(2282) = 2$

$f(1886) = ?$

for some function $f$?

If that is what you mean, then there are an infinite number of functions that satisfy that. Without context, it is impossible to know which answer is correct.

- Mar 25th 2014, 06:08 PM #3

- Joined
- Mar 2014
- From
- Oregon
- Posts
- 3

## Re: Need help please

All the info I have, is what I have supplied. I am wondering if it is some kind of sick joke. My boss said if I can solve it, he will give me a prize. I have been trying to find some kind of pattern for hours. I am not even sure the answer is mathematical. It could be a riddle.

- Mar 25th 2014, 06:31 PM #4

- Joined
- Nov 2010
- Posts
- 2,971
- Thanks
- 1141

## Re: Need help please

Here is a pattern:

Those are all equivalences modulo 1. Of course, modulo 1, all integers are equivalent, so you can pick any integer you want to be the answer, and it will fit that pattern. If your boss doesn't like it, he/she should have given you more specifications.

- Mar 25th 2014, 07:12 PM #5

- Joined
- Nov 2013
- From
- California
- Posts
- 5,880
- Thanks
- 2473

## Re: Need help please

If there's any lesson to be learned from this problem it's that you shouldn't spend a lot of time essentially guessing at answers to a totally under specified problem.

The answer could be anything. It could be the last digit of the weight of that many argon atoms expressed in base 13. Who knows.

- Mar 25th 2014, 07:47 PM #6

- Joined
- Nov 2010
- Posts
- 2,971
- Thanks
- 1141

## Re: Need help please

By the way, in case you don't understand the pattern I pointed out, if two integers, $a$ and $b$, are equivalent modulo some integer $n$, we write:

$a \equiv b \pmod{n}$

That means $\dfrac{a-b}{n}$ is an integer.

Here is an example:

$a \equiv b \pmod{2}$ if and only if both $a$ and $b$ are even or if both are odd.

So, $7746 \equiv 1 \pmod{1}$ means $\dfrac{7746-1}{1}$ is an integer. Because the difference of any two integers is an integer, and any integer divided by 1 is still an integer, $a \equiv b \pmod{1}$ is true for any integers $a$ and $b$. Hence the "pattern" I pointed out for your list of equations.

- Mar 26th 2014, 12:24 AM #7

- Joined
- Mar 2014
- From
- Oregon
- Posts
- 3

## Re: Need help please

I don't know how to edit my original post, but I made a mistake. 6003=3, not 1. My boss gave me the answer to this stupid brain teaser. The answer is 5. The pattern is you count the circles in each group of numbers. An 8 for example has two circles and a 6 has one; therefore, 1886=5.