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Which of the following integers can be written as both the

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New post 25 Apr 2012, 22:56
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Which of the following integers can be written as both the sum of 5 consecutive odd integers and 7 consecutive odd integers?

A. 49
B. 70
C. 140
D. 215
E. 525

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New post 25 Apr 2012, 23:12
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hfbamafan wrote:
Which of the following integers can be written as both the sum of 5 consecutive odd integers and 7 consecutive odd integers?

A. 49
B. 70
C. 140
D. 215
E. 525


First of all notice that both the sum of 5 consecutive odd integers and the sum of 7 consecutive odd integers must be odd. Eliminate B, and C right away.

Next the sum of 5 consecutive odd integers can be written as: (x-4)+(x-2)+x+(x+2)+(x+4)=5x, so the sum must be odd multiple of 5;

Similarly, the sum of 7 consecutive odd integers can be written as: (y-6)+(y-4)+(y-2)+y+(y+2)+(y+4)+(y+6)=7y, so the sum must be odd multiple of 7;

So, the sum must be odd multiple of 5*7=35. Only option E meets this conditions.

Answer: E.
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New post 16 May 2014, 08:53
Hello,
Bunuel one thing I couldn't understand that how the sum is becoming a multiple of 5 & 7. The required sum = 5x + 7x + something. It is not clear to me. Kindly elaborate.
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New post 16 May 2014, 09:43
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deya wrote:
Hello,
Bunuel one thing I couldn't understand that how the sum is becoming a multiple of 5 & 7. The required sum = 5x + 7x + something. It is not clear to me. Kindly elaborate.


I think that you misunderstood the question.

We should be able to write one of the options as the sum of 5 consecutive odd integers. Also, we should be able to write the same options as the sum of 7 consecutive odd integers.

525 = 101 + 103 + 105 + 107 + 109.
525 = 69 + 71 + 73 + 75 + 77 + 79 + 81.

As for 5x and 7y:

The sum of 5 consecutive odd integers can be written as: (x-4)+(x-2)+x+(x+2)+(x+4)=5x --> 5x IS a multiple of 5.
The sum of 7 consecutive odd integers can be written as: (y-6)+(y-4)+(y-2)+y+(y+2)+(y+4)+(y+6)=7y --> 7y IS a multiple of 7.

Thus the integer which is both the sum of 5 consecutive odd integers AND the sum of 7 consecutive odd integers, must be multiple of both 5 and 7, so a multiple of 35
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New post Updated on: 18 Jun 2014, 18:46
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hfbamafan wrote:
Which of the following integers can be written as both the sum of 5 consecutive odd integers and 7 consecutive odd integers?

A. 49
B. 70
C. 140
D. 215
E. 525


This is how I approached the problem.

Sum of 5 consecutive numbers: x + (x + 2) + (x + 4) + (x + 6) + (x + 8) = 5x + 20 --- (1)
Sum of 7 consecutive numbers: y + (y + 2) + (y + 4) + (y + 6) + (y + 8) + (y + 10) + (y + 12) = 7y + 42 --- (2)

Then, equate each of the answer choices with the 2 equations and find out for which Answer Choice you get an ODD VALUE for both 'x' & 'y'.

* Choices (A), (B) & (D) give non-integer values for either x & y on substitution.

* Be careful of Choice (C), as the integer values are EVEN. Not the right answer!! We need ODD consecutive integers.

5x + 20 = 120 ----> x = 24 !!
7y + 42 = 120 ----> y = 14 !!

* Only Choice (E) gives ODD INTEGER values for both x & y.

5x + 20 = 525 ----> x = 101
7y + 42 = 525 ----> y = 69

Hence, (E) is the right answer.

Originally posted by shaderon on 16 May 2014, 22:20.
Last edited by shaderon on 18 Jun 2014, 18:46, edited 1 time in total.
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New post 18 Jun 2014, 08:52
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hfbamafan wrote:
Which of the following integers can be written as both the sum of 5 consecutive odd integers and 7 consecutive odd integers?

A. 49
B. 70
C. 140
D. 215
E. 525


Similar method, just using brute force.

Let's look at the possible values of the sum of 5 consecutive odd integers:

\(1+3+5+7+9 = 10 + 10 + 5 = 5*5 = 25\)
\(3+5+7+9+11 = 25 - 1 + 11 = 25 + 10 = 35\)
\(-3.......+13 = 35 + 10 = 45.\)

We see that each time, the net effect of subtracted and adding will increase the sum by 10. Thus, our sum must be an odd multiple of 5. Our only options are D & E.

Using the same approach for multiples of 7:

\(1+3+5+7+9+11+13 = 49\)
\(3+5+7+9+11+13+15= 49 - 1 + 15 = 49 + 14 = 63\)
\(5+7+9+11+13+15+17= 64-3+17= 63+14 = 77\)

Thus, our number must also be a(n odd) multiple of 7.

D) 215 is clearly not a multiple of 7. Thus, E) 525 (7*75) is our answer.

Answer: E
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New post 19 Jun 2014, 02:40
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LCM of 5 & 7 = 35

Answer should be divisible by 35

Option A & D are cancelled out

Addition of 5 & 7 consecutive odd's would always return an odd number

Option B & C are cancelled out

Answer = E = 525
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Re: Which of the following integers can be written as both the  [#permalink]

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New post 19 Jun 2014, 04:26
arindamsur wrote:
Why not option B.?

Bunuel wrote:
deya wrote:
Hello,
Bunuel one thing I couldn't understand that how the sum is becoming a multiple of 5 & 7. The required sum = 5x + 7x + something. It is not clear to me. Kindly elaborate.


I think that you misunderstood the question.

We should be able to write one of the options as the sum of 5 consecutive odd integers. Also, we should be able to write the same options as the sum of 7 consecutive odd integers.

525 = 101 + 103 + 105 + 107 + 109.
525 = 69 + 71 + 73 + 75 + 77 + 79 + 81.

As for 5x and 7y:

The sum of 5 consecutive odd integers can be written as: (x-4)+(x-2)+x+(x+2)+(x+4)=5x --> 5x IS a multiple of 5.
The sum of 7 consecutive odd integers can be written as: (y-6)+(y-4)+(y-2)+y+(y+2)+(y+4)+(y+6)=7y --> 7y IS a multiple of 7.

Thus the integer which is both the sum of 5 consecutive odd integers AND the sum of 7 consecutive odd integers, must be multiple of both 5 and 7, so a multiple of 35


Because 70 is not an ODD multiple of 35.
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New post 12 Aug 2018, 13:43
In this question, we need to understand that, sum of five consecutive odd numbers will give you an odd number which should end with 5.
So, we are left with options D&E.
Now, as we can see sum of consecutive seven odd numbers(that number) should at least have a factor as 7, which (D) 215 is not.
So, we are left with (525) E.It is our answer.
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New post 19 Aug 2018, 22:41
Bunuel wrote:
deya wrote:
Hello,
Bunuel one thing I couldn't understand that how the sum is becoming a multiple of 5 & 7. The required sum = 5x + 7x + something. It is not clear to me. Kindly elaborate.


I think that you misunderstood the question.

We should be able to write one of the options as the sum of 5 consecutive odd integers. Also, we should be able to write the same options as the sum of 7 consecutive odd integers.

525 = 101 + 103 + 105 + 107 + 109.
525 = 69 + 71 + 73 + 75 + 77 + 79 + 81.

As for 5x and 7y:

The sum of 5 consecutive odd integers can be written as: (x-4)+(x-2)+x+(x+2)+(x+4)=5x --> 5x IS a multiple of 5.
The sum of 7 consecutive odd integers can be written as: (y-6)+(y-4)+(y-2)+y+(y+2)+(y+4)+(y+6)=7y --> 7y IS a multiple of 7.

Thus the integer which is both the sum of 5 consecutive odd integers AND the sum of 7 consecutive odd integers, must be multiple of both 5 and 7, so a multiple of 35




Hi Bunuel,

I followed the same process as mentioned by you, except instead of taking x and y, I took (2x+1) and (2y+1) as the middle terms to ensure that the nos. are odd.
That gave me equations of the form 10x+1 = 525 and 14x+1 = 525, neither of which result in a whole no.
Could you please suggest what's going wrong with this logic?
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New post 19 Aug 2018, 23:52
Shruti0805 wrote:
Bunuel wrote:
deya wrote:
Hello,
Bunuel one thing I couldn't understand that how the sum is becoming a multiple of 5 & 7. The required sum = 5x + 7x + something. It is not clear to me. Kindly elaborate.


I think that you misunderstood the question.

We should be able to write one of the options as the sum of 5 consecutive odd integers. Also, we should be able to write the same options as the sum of 7 consecutive odd integers.

525 = 101 + 103 + 105 + 107 + 109.
525 = 69 + 71 + 73 + 75 + 77 + 79 + 81.

As for 5x and 7y:

The sum of 5 consecutive odd integers can be written as: (x-4)+(x-2)+x+(x+2)+(x+4)=5x --> 5x IS a multiple of 5.
The sum of 7 consecutive odd integers can be written as: (y-6)+(y-4)+(y-2)+y+(y+2)+(y+4)+(y+6)=7y --> 7y IS a multiple of 7.

Thus the integer which is both the sum of 5 consecutive odd integers AND the sum of 7 consecutive odd integers, must be multiple of both 5 and 7, so a multiple of 35




Hi Bunuel,

I followed the same process as mentioned by you, except instead of taking x and y, I took (2x+1) and (2y+1) as the middle terms to ensure that the nos. are odd.
That gave me equations of the form 10x+1 = 525 and 14x+1 = 525, neither of which result in a whole no.
Could you please suggest what's going wrong with this logic?


(2x - 3) + (2x - 1) + (2x +1) + (2x + 3) + (2x + 5) = 525 --> x = 52 --> 2x + 1 = 105
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Re: Which of the following integers can be written as both the  [#permalink]

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New post 07 Oct 2018, 04:15
hfbamafan wrote:
Which of the following integers can be written as both the sum of 5 consecutive odd integers and 7 consecutive odd integers?

A. 49
B. 70
C. 140
D. 215
E. 525


For any evenly spaced set:
sum = (number)(median)

We can PLUG IN THE ANSWERS, which represent the sum.
Since the number of integers can be 5 or 7, the sum must be a multiple of both 5 and 7.
Eliminate A, which is not divisible by 5.
Eliminate D, which is not divisible by 7.

The equation in blue can be rephrased as follows:
\(median = \frac{sum}{number}\)
Since all of the integers must be odd, the correct answer must yield an odd value for the median.

If B is the sum of 7 consecutive odd integers, we get:
\(median = \frac{sum}{number} = \frac{70}{7} = 10\)
Since the yielded median is not odd, eliminate B.
If D is the sum of 7 consecutive odd integers, we get:
\(median = \frac{sum}{number} = \frac{140}{7} = 20\)
Since the yielded median is not odd, eliminate D.


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New post 07 Oct 2018, 07:53
GMATGuruNY wrote:
hfbamafan wrote:
Which of the following integers can be written as both the sum of 5 consecutive odd integers and 7 consecutive odd integers?

A. 49
B. 70
C. 140
D. 215
E. 525


For any evenly spaced set:
sum = (number)(median)

We can PLUG IN THE ANSWERS, which represent the sum.
Since the number of integers can be 5 or 7, the sum must be a multiple of both 5 and 7.
Eliminate A, which is not divisible by 5.
Eliminate D, which is not divisible by 7.

The equation in blue can be rephrased as follows:
\(median = \frac{sum}{number}\)
Since all of the integers must be odd, the correct answer must yield an odd value for the median.

If B is the sum of 7 consecutive odd integers, we get:
\(median = \frac{sum}{number} = \frac{70}{7} = 10\)
Since the yielded median is not odd, eliminate B.
If D is the sum of 7 consecutive odd integers, we get:
\(median = \frac{sum}{number} = \frac{140}{7} = 20\)
Since the yielded median is not odd, eliminate D.



hello GMATGuruNY

what is this fomula used for ? sum = (number)(median)

btw you have 1801 followers (including me) :lol: please correct your automatic signature. thank you. :)
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New post 07 Oct 2018, 08:10
hfbamafan wrote:
Which of the following integers can be written as both the sum of 5 consecutive odd integers and 7 consecutive odd integers?

A. 49
B. 70
C. 140
D. 215
E. 525

\(?\,\,\,:\,\,\,\,\sum\nolimits_5 {\,{\rm{consecutive}}\,\,{\rm{odd}}\,\,{\rm{integers}}\,\,\,\, = \,} \,\,\,\,\sum\nolimits_7 {\,\,{\rm{consecutive}}\,\,{\rm{odd}}\,\,{\rm{integers}}}\)

> 5 consecutive odd integers are members of a finite arithmetic sequence, hence the middle term (one of them) is an odd integer and equal to their average.

(A) average 49/5 is not an integer, hence cannot be the middle term. Refuted.
(B) average 70/5 = 14 is not odd, hence cannot be the middle term. Refuted.
(C) average 140/5 = 28 is not odd, hence cannot be the middle term. Refuted.

> 7 consecutive odd integers are members of a finite arithmetic sequence, hence the middle term (one of them) is an odd integer and equal to their average.

(D) average 215/7 is not a multiple of 7 (210 is), hence cannot be the middle term. Refuted.

(E) is the answer by exclusion.


This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.

P.S.: the middle term is the median, of course.
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Re: Which of the following integers can be written as both the  [#permalink]

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New post 10 Oct 2018, 03:24
dave13 wrote:
hello GMATGuruNY

what is this formula used for ? sum = (number)(median)


The formula holds true for any EVENLY SPACED SET.
An evenly spaced set -- also known as an arithmetic sequence - is a set in which the difference between successive terms is constant.
For example:
{1, 4, 7, 10, 13}
Here:
The difference between successive terms = 3.
The number of terms = 5.
The median = 7.
Sum = (number of terms)(median) = 5*7 = 35.
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New post 12 Oct 2018, 08:03
hfbamafan wrote:
Which of the following integers can be written as both the sum of 5 consecutive odd integers and 7 consecutive odd integers?

A. 49
B. 70
C. 140
D. 215
E. 525


If a number can be written as both the sum of 5 consecutive odd integers and 7 consecutive odd integers, then it’s a multiple of both 5 and 7. This eliminates choice A, 49 (since it’s only a multiple of 7) and choice D, 215 (since it’s only a multiple of 5). So we are left with 70, 140 and 525 to consider.

Not only must the sum be both a multiple of 5 and 7, but the sum also has to be odd (since the sum of five odd numbers must also be odd). This eliminates answer choices B and C and we are left with E.

Answer: E
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New post 14 Apr 2019, 21:13
Bunuel wrote:
hfbamafan wrote:
Which of the following integers can be written as both the sum of 5 consecutive odd integers and 7 consecutive odd integers?

A. 49
B. 70
C. 140
D. 215
E. 525


First of all notice that both the sum of 5 consecutive odd integers and the sum of 7 consecutive odd integers must be odd. Eliminate B, and C right away.

Next the sum of 5 consecutive odd integers can be written as: (x-4)+(x-2)+x+(x+2)+(x+4)=5x, so the sum must be odd multiple of 5;

Similarly, the sum of 7 consecutive odd integers can be written as: (y-6)+(y-4)+(y-2)+y+(y+2)+(y+4)+(y+6)=7y, so the sum must be odd multiple of 7;

So, the sum must be odd multiple of 5*7=35. Only option E meets this conditions.

Answer: E.


Is there a rule which states the sum on n conesuctive integers should be a multiple of n?
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New post 15 Apr 2019, 20:48
hfbamafan wrote:
Which of the following integers can be written as both the sum of 5 consecutive odd integers and 7 consecutive odd integers?

A. 49
B. 70
C. 140
D. 215
E. 525


because the sum is a product
of an odd mean times an odd number of terms,
sum must be an odd multiple of 5 and 7, or 35
only 525 fits
E
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New post 19 Apr 2019, 06:49
The rule about the divisibility of n consecutive integers by n depends on the parity of n. For even values of n, the sum of n consecutive integers is never a multiple of n. If n is odd, the sum of n consecutive integers IS a multiple of n. Suppose the smallest of your consecutive integers is x. Then, the n consecutive integers in your list will be x, x + 1, x + 2, ... , x + (n - 1). The sum is:

x + x + 1 + x + 2 + … + x + (n - 1) = nx + 1 + 2 + … + (n - 1) = nx + (n - 1)*n*(1/2).

As you can see, if n is odd, n - 1 is even and (n - 1)*n*(1/2) is a multiple of n (because (n - 1)*(1/2) is an integer. W

hen n is even, on the other hand, (n - 1)*n*(1/2) is never a multiple of n (because n - 1 is odd and the factor 1/2 cancels one of the factors of 2 in n, resulting in a number that is never a multiple of n.

The sum of consecutive odd integers and consecutive even integers is a different story.

Suppose you have n consecutive odd (or even) integers. Denote the smallest of them by x. Then, the integers are x, x + 2, x + 4, ... , x + 2(n - 1). The sum is:

x + x + 2 + x + 4 + ... + x + 2(n - 1) = nx + 2 + 4 + ... + 2(n-1) = nx + 2(1 + 2 + ... + n -1) = nx + 2*(n - 1)*n*(1/2) = nx + (n - 1)*n

As you can see, the sum, nx + n(n - 1) is always a multiple of n. Notice that it does not matter whether x is even or odd; in either case, the difference between consecutive terms is 2.
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