Question

1. How many sets of two or more consecutive positive integers have a sum of 15?​

Answer 1

There are three three possible sets.

Step-by-step explanation:

• If the sum of n consecutive positive integers starting with a equals 15, then 1/2n(2a+n−1)=15
• n(2a+n−1)=30 with a≥1 and n≥2.
• Thus, n∣30, and so (n,2a+n−1)∈{(2,15),(3,10),(5,6)} (since 2≤ n ≤2a +n−1)
• These give (n,a)∈{(2,7),(3,4),(5,1)}

The other three possible sets are

• 7+8= 15
• 4+5+6=15
• 1+2+3+4+5 = 15

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The members of set A are the integer solutions of the inequality 2x−5≤11 and the members of set B are the integer solutions of the inequality −2x+7≤−9. What is one member of the intersection of A and B?

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