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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

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.

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.

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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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.

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.

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.

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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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

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.

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

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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.

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

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24 Mar 2017, 22:20

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

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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 D

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

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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.
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