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

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.

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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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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Good question.

Thanks for the explanation.

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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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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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

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
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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

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

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

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.

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.

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


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

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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