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 envel opes you could buy so that there is one case for each envelope with none left over
1 answer:
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.
You might be interested in
Answer:
Answer:
y
=18
Explanation
Step-by-step explanation:
<span>B. $49,036 hope helps APEX </span>
Answer:
I aint never seen 2 pretty best friends ;)
Step-by-step explanation:
The constant would be 35 because a variable added onto it would make it a
Answer:
10.083
since 1/12 is .0833333333 and so on, it’s shortened
Step-by-step explanation: