mtb0minime
minimember
+2,418|7135

There are a certain number of things. When I count them by 3's, there are 2 left over. When I count them by 5's, there are 3 left over. When I count them by 7's, there are 2 left over.

How many things are there?


He says there's a systematic way to solve it using "modular arithmetic". I have no idea what that is, but I did a combination of trial-and-error and writing out possibilities based on each hint and finished within a minute or two.

Can you solve it? First to get it gets an one krmaz.
Ollie
Formerly known as Larkin
+215|6464|Halifax, West Yorkshire
I'd rather re-enact "the pounder" with a peice of 2x4 wrapped in barbed-wire than solve that.

Can't wait to see who gets it though.
Jenspm
penis
+1,716|7212|St. Andrews / Oslo

Easy.

(x-2)/3=whole number
(x-2)/7=whole number
(x-3)/5=whole number

Find something thAt can be divided by both 3 and 7: 21

21+2 =23

23-3=20 which can be divided by 5
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mtb0minime
minimember
+2,418|7135

Yay!


Well this topic certainly had a long and fruitful life


Dare I delete?
some_random_panda
Flamesuit essential
+454|6871

I remember actual modulus.

It's like if you divide something by something, you get a remainder.  Now if you add that something to another number with another remainder when divided as above, you add the remainders or something.  Then you can multiply your remainders, etc.
Jenspm
penis
+1,716|7212|St. Andrews / Oslo

some_random_panda wrote:

I remember actual modulus.

It's like if you divide something by something, you get a remainder.  Now if you add that something to another number with another remainder when divided as above, you add the remainders or something.  Then you can multiply your remainders, etc.
That made no sence at all
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