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

Question

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

walkerme13 · Accepted Answer

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.

Bunuel · Answer

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.

Bunuel · Answer

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.

tfincham86 · Answer

Good question.

Thanks for the explanation.

walkerme13 · Answer

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.

Bunuel · Answer

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.

yezz · Answer

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.

Bunuel · Answer

Nothing wrong: x=11 --> x-1=10 (the lowest of the tree integers). The same result as I have in my post.

nelson1972 · Answer

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

EMPOWERgmatRichC · Answer

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.

GMAT assassins aren't born, they're made,
Rich

Temurkhon · Answer

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)

D

mvictor · Answer

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

gracie · Answer

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
D

RMD007 · Answer

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.

chikki420 · Answer

This question clearly said that " the sum of n consecutive positive numbers is 33 ", The n mentioned in the question is the number of integers summed, which is given by the formula S = 2a + (n - 1)d.

where "a" is first term , n is number of integers and d is common difference.

if we use this formula and put S = 33 , n = 3, 6, 11 and d = 1 (because they are consecutive), we get a = 10, 3, -2.

clearly 1 & 2 survived the test and hence Option D is correct.

GMATGuruNY · Answer

For any set of consecutive integers:
 median = \frac{sum}{count}

I: 3 integers
 median = \frac{33}{3} = 11  --> 10, 11, 12

II: 6 integers
 median = \frac{33}{6} = 5.5  --> 3, 4, 5, 6, 7, 8

III: 11 integers
 median = \frac{33}{11} = 3  --> not viable, since some of the integers below the median will be negative

D

chikki420 · Answer

Why to use median concept for the question which belongs to Arithmetic progression, besides the term MEDIAN isnt even mentioned in the question.

GMATGuruNY · Answer

The median of consecutive integers must be either an integer or a value halfway between two integers (-1.5, 0.5, 10.5, etc.).
Here, calculating the median is an efficient way to determine whether a viable set of consecutive integers can be yielded by Statements I, II and III.
A valid statement must satisfy two conditions:
1. The median must be either an integer or a value halfway between two integers
2. The integers surrounding the median must all be positive

As shown in my earlier solution:
Statements I, II and III all satisfy the first condition.
Only Statements I and II satisfy the second condition.
Thus, the correct answer is D.

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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

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