SOLUTION: the integer 33 is to be expressed as a sum of n consecutive positive integers. The value of n could be which of the following ? I. 2 II. 3 III. 6 (A) I only (B)

Algebra ->  Problems-with-consecutive-odd-even-integers -> SOLUTION: the integer 33 is to be expressed as a sum of n consecutive positive integers. The value of n could be which of the following ? I. 2 II. 3 III. 6 (A) I only (B)      Log On


   



Question 312732: the integer 33 is to be expressed as a sum of n consecutive positive integers. The value of n could be which of the following ?


I. 2
II. 3
III. 6
(A) I only
(B) II only
(C) I and II only
(D) I and III only
(E) I, II, and III



Answer by Edwin McCravy(20060) About Me  (Show Source):
You can put this solution on YOUR website!

The sum of consecutive integers is an arithmetic series
with d=1 and first term a%5B1%5D
 
S%5Bn%5D=%282a%5B1%5D+%2B+%28n-1%29%2Ad%29%28n%2F2%29
 
Substituting d=1 and S%5Bn%5D=33
 
33=%282a%5B1%5D+%2B+%28n-1%29%2A1%29%28n%2F2%29
 
Multiplying through by 2
 
66=%282a%5B1%5D%2Bn-1%29n
 
66=2a%5B1%5Dn%2Bn%5E2-n
 
0=n%5E2%2B%282a%5B1%5D-1%29n-66
 
n%5E2%2B%282a%5B1%5D-1%29n-66=0
 
There are only 4 possible factorizations possible for that
so that the middle term's coefficient will be positive: 
 
1.  (n+66)(n-1)=0 which has middle term 65n
 
So n=1.  We can't have just one "consecutive" integer, for
there has to be at least two for each to be "consecutive" to.
 
2.  (n+33)(n-2)=0 which has middle term 31n

So 2a%5B1%5D-1=31
   2a%5B1%5D=32
   a%5B1%5D=16
 
and n=2, and the 2 consecutive integers are 16+17=33

3.  (n+22)(n-3)=0 which has middle term 20n

So 2a%5B1%5D-1=19
   2a%5B1%5D=20
   a%5B1%5D=10

and n=3, and the 3 consecutive integers are 10+11+12=33


4.  (n+11)(n-6)=0 which has middle term 5n

So 2a%5B1%5D-1=5
   2a%5B1%5D=6
   a%5B1%5D=3

and n=6, and the 6 consecutive integers are 3+4+5+6+7+8=33

So there are three possibilities for n, 2, 3, and 6.

That's all three I, II and III which is choice (E)

Edwin