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consecutive integers

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by vscid » Sat Feb 20, 2010 12:25 pm
The sum of n consecutive positive integers is 45. What is the value of n?

(1) n is even

(2) n < 9

How do you solve this quickly?
Last edited by vscid on Sat Feb 20, 2010 12:39 pm, edited 1 time in total.
The GMAT is indeed adaptable. Whenever I answer RC, it proficiently 'adapts' itself to mark my 'right' answer 'wrong'.
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Source: — Data Sufficiency |
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by thephoenix » Sat Feb 20, 2010 12:35 pm
vscid wrote:The sum of n consecutive positive integers is 45. What is the value of n?

(1) n is even

(2) n < 9
(1) n=2 --> 22+23=45, n=4 --> n=6 x1+(x1+1)+(x1+2)+(x1+3)+(x1+4)+(x1+5)=45 x1=5. At least two options for n. Not sufficient.
(2) n<9 same thing not sufficient.
(1)+(2) No new info. Not sufficient.

hence E
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by ajith » Sat Feb 20, 2010 12:39 pm
vscid wrote:The sum of n consecutive positive integers is 45. What is the value of n?

(1) n is even

(2) n < 9
k+1, k+2... , k+n be the numbers

nk+ n(n+1)/2 =45

n(2k+n+1) =90

1) n is even

Insufficient

The numbers will be 22, 23 and n =2

if n =4 the sum will be even

if n =6; 5+6+7+8+9+10 =45

2) Insufficient (same examples)

Combined also insufficient (same examples)
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by shashank.ism » Sat Feb 20, 2010 12:40 pm
vscid wrote:The sum of n consecutive positive integers is 45. What is the value of n?

(1) n is even

(2) n < 9
let say the no. starts from p
so p+(p+1)+(p+2)+.....(p+n-1) = 45
now we start with the combo from 1st
1+2+3+4+5+6+7+8+9 =45

45 45/2= 22.5 so 22+23 will do
45/3 = 15 so 14+15+16 will do
45/4 =11... 10+11+12+13=46 not possible
45/5 = 9 7+8 +9+10 +11 will do
45/6 = 7... 5+6+7+8+9+10 will do
......

so here we see
st.1 : n is even a lot of combo can be there ..22+23, 5+6+7+8+9+10 etc.. not sufficient
st.2 n<9 there can be many combination....as above...... not sufficient
combined also not sufficient

Ans E
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by vscid » Mon Feb 22, 2010 2:07 pm
shashank.ism wrote:
vscid wrote:The sum of n consecutive positive integers is 45. What is the value of n?

(1) n is even

(2) n < 9
let say the no. starts from p
so p+(p+1)+(p+2)+.....(p+n-1) = 45
now we start with the combo from 1st
1+2+3+4+5+6+7+8+9 =45

45 45/2= 22.5 so 22+23 will do
45/3 = 15 so 14+15+16 will do
45/4 =11... 10+11+12+13=46 not possible
45/5 = 9 7+8 +9+10 +11 will do
45/6 = 7... 5+6+7+8+9+10 will do
......

so here we see
st.1 : n is even a lot of combo can be there ..22+23, 5+6+7+8+9+10 etc.. not sufficient
st.2 n<9 there can be many combination....as above...... not sufficient
combined also not sufficient

Ans E
Shashank, thats a nice approach.
OA E.
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by lunarpower » Tue Feb 23, 2010 6:23 am
shashank.ism wrote:
vscid wrote:The sum of n consecutive positive integers is 45. What is the value of n?

(1) n is even

(2) n < 9
let say the no. starts from p
so p+(p+1)+(p+2)+.....(p+n-1) = 45
now we start with the combo from 1st
1+2+3+4+5+6+7+8+9 =45

45 45/2= 22.5 so 22+23 will do
45/3 = 15 so 14+15+16 will do
45/4 =11... 10+11+12+13=46 not possible
45/5 = 9 7+8 +9+10 +11 will do
45/6 = 7... 5+6+7+8+9+10 will do
......

so here we see
st.1 : n is even a lot of combo can be there ..22+23, 5+6+7+8+9+10 etc.. not sufficient
st.2 n<9 there can be many combination....as above...... not sufficient
combined also not sufficient

Ans E
beautifully done.

this problem, by the way, typifies a very common theme on the gmat math section: sometimes, MAKING AN EXHAUSTIVE LIST is instrumental in solving a problem.
there's really no way to solve this problem other than LISTING all of these possibilities.

if you're a "stubborn algebra fiend", take note - there are some problems where you HAVE to just list possibilities.
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by STEVEN SPIELBERG » Sat Mar 13, 2010 11:16 am
E . Good approach !!
I want to win an OSCAR on the GMAT !!!
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