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# A page is torn from a novel. The sum of the remaining page numbers is

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Math Expert
Joined: 02 Sep 2009
Posts: 65187
A page is torn from a novel. The sum of the remaining page numbers is  [#permalink]

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14 Apr 2020, 08:08
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Difficulty:

75% (hard)

Question Stats:

29% (02:02) correct 71% (02:31) wrong based on 51 sessions

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A page is torn from a novel. The sum of the remaining page numbers is 10,000. What could be the page-numbers on the torn page ?

I. 76 and 77
II. 33 and 34
III. 5 and 6.

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

Are You Up For the Challenge: 700 Level Questions

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Re: A page is torn from a novel. The sum of the remaining page numbers is  [#permalink]

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14 Apr 2020, 08:23
Bunuel wrote:
A page is torn from a novel. The sum of the remaining page numbers is 10,000. What could be the page-numbers on the torn page ?

I. 76 and 77
II. 33 and 34
III. 5 and 6.

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

Are You Up For the Challenge: 700 Level Questions

CONCEPT: Sum of first n consecutive integers, $$1+2+3....+n = (\frac{1}{2})*n(n+1)$$

Here the sum should be close to 10,000 and greater than 10,000

Lets calculate (1/2)*n(n+1) = 10000
n*(n+1) = 20000
i.e.n ≈ 141

141*142 = 20022

i.e. (1/2)*141*142 = 10011

So ew are short of 11

OR

We are short of 142(next page)+11 = 153

I. 76 and 77 Sum of 76 and 77 = 153 POSSIBLE
II. 33 and 34
III. 5 and 6. Sum of 5 and 6 = 11 POSSIBLE

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Re: A page is torn from a novel. The sum of the remaining page numbers is  [#permalink]

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14 Apr 2020, 12:13
GMATinsight wrote:
Bunuel wrote:
A page is torn from a novel. The sum of the remaining page numbers is 10,000. What could be the page-numbers on the torn page ?

I. 76 and 77
II. 33 and 34
III. 5 and 6.

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

Are You Up For the Challenge: 700 Level Questions

CONCEPT: Sum of first n consecutive integers, $$1+2+3....+n = (\frac{1}{2})*n(n+1)$$

Here the sum should be close to 10,000 and greater than 10,000

Lets calculate (1/2)*n(n+1) = 10000
n*(n+1) = 20000
i.e.n ≈ 141

141*142 = 20022

i.e. (1/2)*141*142 = 10011

So ew are short of 11

OR

We are short of 142(next page)+11 = 153

I. 76 and 77 Sum of 76 and 77 = 153 POSSIBLE
II. 33 and 34
III. 5 and 6. Sum of 5 and 6 = 11 POSSIBLE

Hi Can you Pls explain how did you get 142 + 11= 153. Thanks in advance.
Senior Manager
Joined: 18 Dec 2017
Posts: 300
Re: A page is torn from a novel. The sum of the remaining page numbers is  [#permalink]

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14 Apr 2020, 19:32
GMATinsight wrote:
Bunuel wrote:
A page is torn from a novel. The sum of the remaining page numbers is 10,000. What could be the page-numbers on the torn page ?

I. 76 and 77
II. 33 and 34
III. 5 and 6.

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

Are You Up For the Challenge: 700 Level Questions

CONCEPT: Sum of first n consecutive integers, $$1+2+3....+n = (\frac{1}{2})*n(n+1)$$

Here the sum should be close to 10,000 and greater than 10,000

Lets calculate (1/2)*n(n+1) = 10000
n*(n+1) = 20000
i.e.n ≈ 141

141*142 = 20022

i.e. (1/2)*141*142 = 10011

So ew are short of 11

OR

We are short of 142(next page)+11 = 153

I. 76 and 77 Sum of 76 and 77 = 153 POSSIBLE
II. 33 and 34
III. 5 and 6. Sum of 5 and 6 = 11 POSSIBLE

Hi Can you Pls explain how did you get 142 + 11= 153. Thanks in advance.

Suppose instead of 141 pages in the book, if there were 142 pages then sum of the pages would have been 10153 then extra sum would have been 153
Which is 77 +76 =153
So this is also the possibility

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Re: A page is torn from a novel. The sum of the remaining page numbers is  [#permalink]

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14 Apr 2020, 23:29
1
GMATinsight wrote:
Bunuel wrote:
A page is torn from a novel. The sum of the remaining page numbers is 10,000. What could be the page-numbers on the torn page ?

I. 76 and 77
II. 33 and 34
III. 5 and 6.

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

Are You Up For the Challenge: 700 Level Questions

CONCEPT: Sum of first n consecutive integers, $$1+2+3....+n = (\frac{1}{2})*n(n+1)$$

Here the sum should be close to 10,000 and greater than 10,000

Lets calculate (1/2)*n(n+1) = 10000
n*(n+1) = 20000
i.e.n ≈ 141

141*142 = 20022

i.e. (1/2)*141*142 = 10011

So we are short of 11

OR

We are short of 142(next page)+11 = 153

I. 76 and 77 Sum of 76 and 77 = 153 POSSIBLE
II. 33 and 34
III. 5 and 6. Sum of 5 and 6 = 11 POSSIBLE

Hi Can you Pls explain how did you get 142 + 11= 153. Thanks in advance.

Sum of 1 to 141 is (1/2)*141*142 = 10011 so the sum of page number should be 11 more i.e. pages torn must have sum of page numbers 11

But

other possibilities is that there were 142 pages and in that case the missing sum would be 11+142(next page number) = 153

I hope this help!!!
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Re: A page is torn from a novel. The sum of the remaining page numbers is  [#permalink]

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15 Apr 2020, 06:29
thanks! Got it! Basically x=141 or 142. I missed it.
Thanks for the explanation.
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Joined: 09 Oct 2019
Posts: 6
Re: A page is torn from a novel. The sum of the remaining page numbers is  [#permalink]

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16 Apr 2020, 02:46
I think it's A because a book/novel can only have even pages
So for 141 pages is no chance
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Re: A page is torn from a novel. The sum of the remaining page numbers is  [#permalink]

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16 Apr 2020, 06:10
rudi0setyawan wrote:
I think it's A because a book/novel can only have even pages
So for 141 pages is no chance

rudi0setyawan

What makes you think that a book/novel can't have odd pages.

It's only the page number which can very well be even or odd

The numbering is done in many ways in books. A few pages in beginning in Roman count where the introduction is mentioned, the dedication letter is mentiones.

Towards the end there may be some advertise and all and the numbers mentioned on pages can well be Odd number too
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Re: A page is torn from a novel. The sum of the remaining page numbers is  [#permalink]

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17 Apr 2020, 10:21
2
1
Bunuel wrote:
A page is torn from a novel. The sum of the remaining page numbers is 10,000. What could be the page-numbers on the torn page ?

I. 76 and 77
II. 33 and 34
III. 5 and 6.

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

Are You Up For the Challenge: 700 Level Questions

Recall that the sum of the consecutive integers from 1 to n is given by n(n + 1)/2. Let’s analyze each Roman numeral:

I: 76 and 77

If the page numbers on the torn page was 76 and 77, then the sum of the pages of the novel was 10,000 + 76 + 77 = 10,153. To determine whether 10,153 is the sum of consecutive integers, let’s solve n(n + 1)/2 = 10,153:

n(n + 1)/2 = 10153

n^2 + n = 20306

n^2 + n - 20306 = 0

(n + 143)(n - 142) = 0

n = -143 or n = 142

We see that 10,153 is the sum of the integers from 1 to 142; therefore it is possible that the page numbers on the torn page were 76 and 77.

II: 33 and 34

If the page numbers on the torn page was 33 and 34, then the sum of the pages of the novel was 10,000 + 33 + 34 = 10,067. To determine whether 10,067 is the sum of consecutive integers, let’s solve n(n + 1)/2 = 10,067:

n(n+1)/2 = 10067

n^2 + n = 20134

n^2 + n - 20134 = 0

This equation has no integer roots. Thus, the page numbers on the torn page cannot be 33 and 34.

III: 5 and 6

If the page numbers on the torn page was 5 and 6, then the sum of the pages of the novel was 10,000 + 5 + 6 = 10,011. To determine whether 10,011 is the sum of consecutive integers, let’s solve n(n + 1)/2 = 10,011:

n(n+1)/2 = 10011

n^2 + n = 20022

n^2 + n - 20022 = 0

(n + 142)(n - 141) = 0

We see that 10,011 is the sum of the integers from 1 to 141; therefore it is possible that the page numbers on the torn page were 5 and 6.

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A page is torn from a novel. The sum of the remaining page numbers is  [#permalink]

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09 Jul 2020, 12:31
Recall that the sum of the consecutive integers from 1 to n is given by n(n + 1)/2. Let’s analyze each Roman numeral:

I: 76 and 77

If the page numbers on the torn page was 76 and 77, then the sum of the pages of the novel was 10,000 + 76 + 77 = 10,153. To determine whether 10,153 is the sum of consecutive integers, let’s solve n(n + 1)/2 = 10,153:

n(n + 1)/2 = 10153

n^2 + n = 20306

n^2 + n - 20306 = 0

(n + 143)(n - 142) = 0

n = -143 or n = 142

We see that 10,153 is the sum of the integers from 1 to 142; therefore it is possible that the page numbers on the torn page were 76 and 77.

II: 33 and 34

If the page numbers on the torn page was 33 and 34, then the sum of the pages of the novel was 10,000 + 33 + 34 = 10,067. To determine whether 10,067 is the sum of consecutive integers, let’s solve n(n + 1)/2 = 10,067:

n(n+1)/2 = 10067

n^2 + n = 20134

n^2 + n - 20134 = 0

This equation has no integer roots. Thus, the page numbers on the torn page cannot be 33 and 34.

III: 5 and 6

If the page numbers on the torn page was 5 and 6, then the sum of the pages of the novel was 10,000 + 5 + 6 = 10,011. To determine whether 10,011 is the sum of consecutive integers, let’s solve n(n + 1)/2 = 10,011:

n(n+1)/2 = 10011

n^2 + n = 20022

n^2 + n - 20022 = 0

(n + 142)(n - 141) = 0

We see that 10,011 is the sum of the integers from 1 to 141; therefore it is possible that the page numbers on the torn page were 5 and 6.

[/quote]

Hi ScottTargetTestPrep,

Could you please tell how to split the middle term for such equations (n^2 + n - 20306 = 0
) in a quicker manner as I am finding difficulty in solving them.

Thanks!
A page is torn from a novel. The sum of the remaining page numbers is   [#permalink] 09 Jul 2020, 12:31