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If the sum of n consecutive positive integers is 33, what of

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If the sum of n consecutive positive integers is 33, what of [#permalink]

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If the sum of n consecutive positive integers is 33, what of the following could be the value of n?

I. 3
II. 6
III. 11

A. I only
B. II only
C. III only
D. I and II
E. I, II and III
[Reveal] Spoiler: OA
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Re: #37 sum of consecutive integers [#permalink]

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aalriy wrote:
If the sum of \(n\) consecutive positive integers is 33, what of the following could be the value of \(n\)?
I. 3
II. 6
III. 11

A. I only
B. II only
C. III only
D. I and II
E. I, II and III


You can solve this question with some formulas but trial and error will give an answer in less than 1 min.

Can n equal to 3? It's easy to find that 10+11+12=33, so yes;
Can n equal to 6? Again it's easy to find that 3+4+5+6+7+8=33, so yes;
Can n equal to 11? If we take the smallest 11 consecutive positive integers: 1, 2, 3, ..., 10, 11 we'll see that 11+10+9+8 is already more than 33, so n can not equal to 11.

Answer: D.
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Re: #37 sum of consecutive integers [#permalink]

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I think I used a better way..

u have 3 options so test for each option

1st option is 3. So n + n+1 + n+2 = 33. so u get 3n + 3 = 33. n = 10. So 3 works.

2nd option is 6. so n + n+1 + n+2 + n+3 + n+4 + n+5 = 33. 6n + 15 + 33. n = 3. So 6 works .

3rd option. We can clearly see through common sense that this option doesnt work as the last 3 least possible consecutive number 9 10 and 11 gives u 30 already.

I think if you are quick it should take less than a minute easily and u can use this standard form for similar problems.
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Re: #37 sum of consecutive integers [#permalink]

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walkerme13 wrote:
I think I used a better way..

u have 3 options so test for each option

1st option is 3. So n + n+1 + n+2 = 33. so u get 3n + 3 = 33. n = 10. So 3 works.

2nd option is 6. so n + n+1 + n+2 + n+3 + n+4 + n+5 = 33. 6n + 15 + 33. n = 3. So 6 works .

3rd option. We can clearly see through common sense that this option doesnt work as the last 3 least possible consecutive number 9 10 and 11 gives u 30 already.

I think if you are quick it should take less than a minute easily and u can use this standard form for similar problems.


Yes, you can use a variable to express sequence of consecutive integers, though it's better to use some other than n as n is already used for # of terms:

x+(x+1)+(x+2)=33 --> x=10;
x+(x+1)+(x+2)+(x+3)+(x+4)+(x+5)=33 --> x=3;
x+(x+1)+(x+2)+...+(x+10)=33 --> x=-2, not possible as we are told that the sequence contains positive integers only.
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Re: #37 sum of consecutive integers [#permalink]

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New post 10 Dec 2010, 09:23
Good question.

Thanks for the explanation.
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Re: #37 sum of consecutive integers [#permalink]

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yes. you are right using x would avoid confusion. But I wonder if this method is good for higher numbers. number of integers more than 15 etc. but i dont think they would ask such questions so we are safe.
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Re: #37 sum of consecutive integers [#permalink]

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New post 10 Dec 2010, 09:43
walkerme13 wrote:
yes. you are right using x would avoid confusion. But I wonder if this method is good for higher numbers. number of integers more than 15 etc. but i dont think they would ask such questions so we are safe.


This method is perfectly OK with higher # of terms. For example: the sum of 25 consecutive numbers is 500, what is the first number?

Consecutive numbers can be expressed not only as x, x+1, x+2, ... but also ..., x-2, x-1, x, x+1, x+2, ...

So, 25 consecutive integers can be expressed as x-12, x-11, ..., x-1, x, x+1, ..., x+12 --> when we add them up we'll have: (x-12)+(x-11)+...+(x-1)+x+(x+1)+...+(x+12)=500 --> 25x=500 --> x=20 --> first # is x-12=8.

Of course you can use AP formulas as well for such kind of questions, check this: math-number-theory-88376.html and this: sequences-progressions-101891.html for more in this topic.

Hope it helps.
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Re: #37 sum of consecutive integers [#permalink]

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New post 25 May 2013, 04:06
Bunuel wrote:
walkerme13 wrote:
I think I used a better way..

u have 3 options so test for each option

1st option is 3. So n + n+1 + n+2 = 33. so u get 3n + 3 = 33. n = 10. So 3 works.

2nd option is 6. so n + n+1 + n+2 + n+3 + n+4 + n+5 = 33. 6n + 15 + 33. n = 3. So 6 works .

3rd option. We can clearly see through common sense that this option doesnt work as the last 3 least possible consecutive number 9 10 and 11 gives u 30 already.

I think if you are quick it should take less than a minute easily and u can use this standard form for similar problems.


Yes, you can use a variable to express sequence of consecutive integers, though it's better to use some other than n as n is already used for # of terms:

x+(x+1)+(x+2)=33 --> x=10;
x+(x+1)+(x+2)+(x+3)+(x+4)+(x+5)=33 --> x=3;
x+(x+1)+(x+2)+...+(x+10)=33 --> x=-2, not possible as we are told that the sequence contains positive integers only.



Hi Bunuel , i used x-1,x,x+1 and it didnt work it gives rise to 3x=33?? where am i going wrong if u plz.
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Re: #37 sum of consecutive integers [#permalink]

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New post 25 May 2013, 04:14
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yezz wrote:
Bunuel wrote:
walkerme13 wrote:
I think I used a better way..

u have 3 options so test for each option

1st option is 3. So n + n+1 + n+2 = 33. so u get 3n + 3 = 33. n = 10. So 3 works.

2nd option is 6. so n + n+1 + n+2 + n+3 + n+4 + n+5 = 33. 6n + 15 + 33. n = 3. So 6 works .

3rd option. We can clearly see through common sense that this option doesnt work as the last 3 least possible consecutive number 9 10 and 11 gives u 30 already.

I think if you are quick it should take less than a minute easily and u can use this standard form for similar problems.


Yes, you can use a variable to express sequence of consecutive integers, though it's better to use some other than n as n is already used for # of terms:

x+(x+1)+(x+2)=33 --> x=10;
x+(x+1)+(x+2)+(x+3)+(x+4)+(x+5)=33 --> x=3;
x+(x+1)+(x+2)+...+(x+10)=33 --> x=-2, not possible as we are told that the sequence contains positive integers only.



Hi Bunuel , i used x-1,x,x+1 and it didnt work it gives rise to 3x=33?? where am i going wrong if u plz.


Nothing wrong: x=11 --> x-1=10 (the lowest of the tree integers). The same result as I have in my post.
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Re: If the sum of n consecutive positive integers is 33, what of [#permalink]

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New post 11 Jan 2015, 11:56
Thank for the posot. What confuses me is that one approach works for some numbers, while other for others. Exa, for 3 and 6 you can use x, x+1, etc and come up with 3x + 3 =33, and 6x + 15= 33, which gives you x=10, x= 3 =. OK.

But what happens with same rule for n=11? there we have to come up with dif approach: none of the approach above seems to work for me ( help me where I may be wrong ).
Using x-5, x-4, for n=11 doesnt work for me, neither the 11 n + N, which gives x<0 as properly indicated by Brunel above This n=11 is the most time consuming option to evaluate.

Is there any more general approach?
Thaks guys for great help!
Nelson
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Re: If the sum of n consecutive positive integers is 33, what of [#permalink]

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New post 11 Jan 2015, 16:59
Hi nelson1972,

You seem comfortable with how the approach proves that N COULD be 3 or 6. The approach will also help you to prove that N CANNOT be 11.

Using the same logic...the 11 terms would be...
X
X+1
X+2
.....
X+10

The sum would be...
11X + 55

With the restriction that the sum of the terms MUST = 33 AND that all 11 terms MUST be POSITIVE INTEGERS.....

Does this equation have a solution:
11X + 55 = 33

11X = - 22
X = -2

However, we're told that the terms MUST be POSITIVE INTEGERS, so there CAN'T be 11 terms (the first 3 terms in THIS sequence would be -2, -1 and 0 which doesn't match the "restrictions"). This work helps to eliminate this option.

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Re: If the sum of n consecutive positive integers is 33, what of [#permalink]

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New post 11 Jan 2015, 23:21
We have consecutive integer, so their sum is the average multiplyied by N. Average=median in consecutive sets

N=3 means that average and median is 11, so set it 10,11,12 and sum is 33 (possible)

N=6 means that median is between 5 and 6 giving by itself 11, so it is 3,4,5,6,7,8 and sum is 33 (possible)

N=11 means that average and median is 3, so set can be only 1,2,3,4,5 or even shorter and never sum is 33
(impossible)

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Re: If the sum of n consecutive positive integers is 33, what of [#permalink]

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New post 01 Jan 2016, 19:12
solved it the way bunuel explained, although took some time to arrive to the answer choice.
3 -> 10+11+12=33.
6 ->3+4+5+6+7+8=33.
11 -> since we are told that we have a set of n consecutive POSITIVE numbers, we can have the smallest possible numbers:
1+2+3+4+5+6+7+8+9+10+11 - the sum is way over 33, it is actually 11*12/2 = 11*6=66.

thus, only I and II works. the answer is D
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Re: If the sum of n consecutive positive integers is 33, what of [#permalink]

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Re: If the sum of n consecutive positive integers is 33, what of [#permalink]

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New post 25 Mar 2017, 12:23
aalriy wrote:
If the sum of n consecutive positive integers is 33, what of the following could be the value of n?

I. 3
II. 6
III. 11

A. I only
B. II only
C. III only
D. I and II
E. I, II and III


one approach is to divide the sum by the choices for n
if both are odd or both even, the quotient gives the middle term: 33/3=11
if odd/even or even/odd, the quotient gives the median: 33/6=5.5
33/11 gives 3, which clearly won't work as either a middle term or median
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If the sum of n consecutive positive integers is 33, what of [#permalink]

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New post 29 Mar 2017, 06:34
Bunuel wrote:
aalriy wrote:
If the sum of \(n\) consecutive positive integers is 33, what of the following could be the value of \(n\)?
I. 3
II. 6
III. 11

A. I only
B. II only
C. III only
D. I and II
E. I, II and III


You can solve this question with some formulas but trial and error will give an answer in less than 1 min.

Can n equal to 3? It's easy to find that 10+11+12=33, so yes;
Can n equal to 6? Again it's easy to find that 3+4+5+6+7+8=33, so yes;
Can n equal to 11? If we take the smallest 11 consecutive positive integers: 1, 2, 3, ..., 10, 11 we'll see that 11+10+9+8 is already more than 33, so n can not equal to 11.

Answer: D.


Another approach:

33 is an odd number.

I. 3 --> even + odd + even = odd. This option can result in sum- 33 an odd number. So this COULD BE the value of N.
II. 6 --> for consecutive integers, obviously we will have 3 even + 3 odd numbers = Odd number ==> Possible.

III. 11 --> being such a large number. Least possible option is that the it will be first 11 integers, but first 11 integers sum up to - \(\frac{(11*12)}{2} = 66\) Not possible.
If the sum of n consecutive positive integers is 33, what of   [#permalink] 29 Mar 2017, 06:34
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