A box contains gold coins. If the coins are equally divided among six friends, four coins are left over. If the coins are equally divided among five friends, three coins are left over. If the box holds the smallest number of coins that meets these two conditions, how many coins are left when equally divided among seven friends?

Question

ganeshie8 · Answer

attachment

taylor<3srin · Answer

If there were two more coins in the box, the number of coins would be divisible by both
6 and 5. The smallest number that is divisible by 6 and 5 is 30, so the smallest possible
number of coins in the box is 28.

ganeshie8 · Answer

very good logic @Taylor<3sRin :)

taylor<3srin · Answer

Thank you
Ganeshire!

taylor<3srin · Answer

I keep mistyping on my enter key, I hate my new keyboard.

ganeshie8 · Answer

lol xD
more complicated problems like these can be solved using 'Chinese Remainder Theorem'... incase you haven't got in touch with this yet, have a look...

ganeshie8 · Answer

here is bit harder problem :-
If eggs are removed from a basket 2, 3, 5 or 7 at a time, 1
egg remains left over. But if eggs are removed 11 at a time, no eggs remain.
What is the least number of eggs that could have been in the basket?

ganeshie8 · Answer

give it a try when free :)

taylor<3srin · Answer

Care to walk me through this one?

taylor<3srin · Answer

Yeah I tried, can't figure it out

ganeshie8 · Answer

its bit easy, but we need to knw few things before attempting this problem

ganeshie8 · Answer

say, there are $x$ eggs in the basket.
dividing by 2 leaves a remainder of 1
dividing by 3 leaves a remainder of 1
dividing by 5 leaves a remainder of 1
dividing by 7 leaves a remainder of 1
dividing by 11 leaves a remainder of 0

so, we need to find $x$ wid above properties

taylor<3srin · Answer

I'm pretty much lost now.

goformit100 · Answer

Thankyou