SOLUTION: ricky has some stickers in a box. if he puts the stickers into bags of 9. he will have no stickers left over. if he puts them into bags of 10, he will have 2 stickers left over. wh

Algebra ->  Divisibility and Prime Numbers -> SOLUTION: ricky has some stickers in a box. if he puts the stickers into bags of 9. he will have no stickers left over. if he puts them into bags of 10, he will have 2 stickers left over. wh      Log On


   



Question 1139840: ricky has some stickers in a box. if he puts the stickers into bags of 9. he will have no stickers left over. if he puts them into bags of 10, he will have 2 stickers left over. what is the smallest possible number of stickers in the box?
Can you please help me with this problem.
Thank u

Answer by Edwin McCravy(20054) About Me  (Show Source):
You can put this solution on YOUR website!
Ricky has some stickers in a box.
 
Let N = number of stickers Ricky has in his box.

If he puts the stickers into bags of 9, he will have no stickers left over.
 
So N is a multiple of 9, so N=9P, for some positive integer P.

If he puts them into bags of 10, he will have 2 stickers left over.
 
So N is 2 more than a multiple of 10, so N=9P=10Q+2, for some positive integer Q.

what is the smallest possible number of stickers in the box?
We write the larger coefficient (in absolute value), 10, in terms of the nearest
multiple of the smaller coefficient (in absolute value), 9. So we write 10 as
9+1

    9P=10Q+2
    9P=(9+1)Q+2
    9P=9Q+Q+2
 9P-9Q=Q+2
9(P-Q)=Q+2 
  P-Q=(Q+2)/9

The right side is positive and will be an integer if Q is 2 less
than a multiple of 9, so Q could be 7,16,25,...  Since N=9P, N
will be smallest when P is smallest, so we take P=7

If Q is 7 then 9P=10Q+2
               9P=10(7)+2
               9P=70+2
               9P=72
                P=8

Since N=9P, N=72

The smallest possible number of stickers in the box is 72.

Edwin