Which of the following integers can be written as both the

Question

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

Bunuel · Accepted Answer

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.

Bunuel · Answer

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

deya · Answer

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.

shaderon · Answer

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.

GSBae · Answer

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

PareshGmat · Answer

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

Bunuel · Answer

Because 70 is not an ODD multiple of 35.

Shruti0805 · Answer

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?

Bunuel · Answer

(2x - 3) + (2x - 1) + (2x +1) + (2x + 3) + (2x + 5) = 525 --> x = 52 --> 2x + 1 =  105

GMATGuruNY · Answer

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.

E

dave13 · Answer

hello GMATGuruNY what is this fomula used for ? sum = (number)(median) btw you have 180 1 followers (including me)

please correct your automatic signature. thank you.

fskilnik · Answer

?\,\,\,:\,\,\,\,\sum
olimits_5 {\,{m{consecutive}}\,\,{m{odd}}\,\,{m{integers}}\,\,\,\, = \,} \,\,\,\,\sum
olimits_7 {\,\,{m{consecutive}}\,\,{m{odd}}\,\,{m{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.

GMATGuruNY · Answer

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.

ScottTargetTestPrep · Answer

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

ScottTargetTestPrep · Answer

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.

BrentGMATPrepNow · Answer

We have consecutive odd integers, and each value is  2 greater  than the value before it

So for example, if we let x = the smallest odd integer in the set of five consecutive odd integers then...
The sum of the 5 integers = x + (x + 2) + (x + 4) + (x + 6) + (x + 8) = 5x + 20 = 5(x + 4)
NOTE: since x is ODD, the sum of the 5 integers = 5(ODD + 4) = 5(some odd integer)
In other words, the sum must be an ODD MULTIPLE OF 5 (e.g., 5, 15, 25, 35, 45, etc)

Check the answer choices....
We can ELIMINATE A, B and C since they are NOT odd multiples of 5

Likewise, if we let k = the smallest odd integer in the set of seven consecutive odd integers then...
The sum of the 7 integers = k + (k + 2) + (k + 4) + (k + 6) + (k + 8) + (k + 10) + (k + 12) = 7k + 42 = 7(k + 6)
NOTE: since k is ODD, the sum of the 7 integers = 7(ODD + 6) = 7(some odd integer)
In other words, the sum must be an ODD MULTIPLE OF 7 (e.g., 7, 21, 35, 49, ... etc)

Check the remaining answer choices....
We can ELIMINATE D since it's not even a multiples of 7

By the process of elimination, the correct answer is E

Cheers,
Brent

Gyrononi · Answer

70, 140 and 525 all are multiple of 35. Am I missing something to eliminate 70 and 140?

Bunuel · Answer

Check the highlighted parts in the solution. 70 and 140 are NOT odd.

Gyrononi · Answer

average of first 90 is 9.5.
Average of last 10 is 10.

so weighted mean is sum of (90 *9.5)/(90+10) and (10*10)/(90+10)
which is 8.55+1 = 9.55

Which of the following integers can be written as both the

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