Dvd cases are sold in packages of 20 padded mailing envelopes or so in packets of 12 what is the least number of cases and envelopes you could buy so that there is one case for each envelope with none left over

The least number of cases and envelopes so that there would be no remainder can be determined by finding their least common multiple. The solution is as follows

     4 | 20  12
       
        |  5     3
       

Multiplying 4*5*3 = 60. The least common multiple is 60. Thus, there should be at least 60 each of the envelopes and the packets.

Number of mailing envelopes = 20x = 60
x = 3 
Number of packets = 12y = 60
y = 5

That would be 3 packs of the envelopes, and 5 packs of the packets.

Aye i really don’t know this but i need more point so i am guessing it’s c. 2.0or you can use other apps to you get the answer