someone help with this homework question?

[blondemafia20]

it's called many many marbles


There is a bucket of marbles.
When 2 people divide them equally there is 1 left over.
When 3 people divide them equally there is 1 left over.
It's the same when 4, 5, or 6 people divide them equally.


But when 7 people divide them equally there are no marbles left over.


what is the least number of marbles in the bucket?




1. is there an even number of marbles in the bucket and why?


2) Is the number of marbles in the bucket a multiple of 3? 4? 5? 6? 7?



I couldnt figure this out and I worked on it for like an hour. It's my cousins 7th grade math homework...and I'm 20! lol!

Please explain how you got the answer so i can explain it to him!

Thanks!

[Ryan L]

i thought of 49 but doesn't fit with 5... then i just keep trying multiples of 7 because when people divide by 7 there is no more left. and 91 doesn't work either... just keep getting multiples of 7 until it all works out.
1.No, because 2 would fit into it and maybe 4 too.
2.No to 3,4,5 and 6 because then it would have to be 1 left over.
Try to find a 1 as the one's digit

[AndyL]

7, 7 is the Answer....

[Nuffie]

1. No
2. No, no, no, no, yes

[Ina]

Least number is 43x7 = 301

Not good at explaining but..

Since we can only divide the number of marbles with 7, then the multiples must be 7. The marbles number when divided with 2,3,4,5 and 6 brings 1 remainder. Remember that the multiples of 5 must have 0 or 5 at the back and the multiples 2,6 and 4 must be an even number.
We can't multiply 7 with 2,3,4,5,6 because it won't bring any remainder, so start with 7.

7x7 =49 ~49-1(the remainder)=48. 48 can't be divided with 5.
7x9 =63 ~It can't be this either, 63-1 =62
7x10 =70 ~It can't be because it's an even number.
7x11 =77 ~77-1=76, can't be divide with 5.

You already know that if we multiply 7 with an even number, the answer is even number. So skip any even number. Go straight to the odd number to multiply with 7.

7x13 =91 ~91-1 =90 , you get a 0 at the end, and you can divide with 5 but, you can't divide them with 4. Notice that only when 7 is multiply with a number which has 3 at the end, it can bring 0 when we minus 1, so the number you have to multipy must be 23,33,43,53.63,73 etc so that we can at least multiply with 5.

7x23 =161 ~161-1 =160, can't be divided with 3.
7x33 =231 ~231-1 =230, can't be divided with 6.
7x43 =301 ~301-1 =300, can be divided with 2,3,4,5 and 6.

1. No
2. It's7.

Conclusion:
Least number of marbles: 301
1) No because any even number when divide with 2 have no remainder. So the answer can't be an even number.
2) The number of marbles in the bucket is the multiples of 7.

*I'm so sorry. I'm really bad at explaining but I do hope you understand a bit. Maybe someone else can explain it better. Good Luck and again, forgive me for my weakness.
*I'm not using calculator, so do double check. Sorry for the trouble.