GMAT Question of the Day: Daily via email | Daily via Instagram New to GMAT Club? Watch this Video

 It is currently 24 Jan 2020, 02:32

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# The sum of four consecutive odd numbers is equal to the sum

Author Message
TAGS:

### Hide Tags

Intern
Joined: 09 Aug 2011
Posts: 7
Location: India
Concentration: Finance, Strategy
GMAT Date: 09-30-2011
WE: Corporate Finance (Energy and Utilities)
The sum of four consecutive odd numbers is equal to the sum  [#permalink]

### Show Tags

Updated on: 25 Sep 2013, 01:36
18
144
00:00

Difficulty:

95% (hard)

Question Stats:

24% (02:58) correct 76% (02:52) wrong based on 1341 sessions

### HideShow timer Statistics

The sum of four consecutive odd numbers is equal to the sum of 3 consecutive even numbers. Given that the middle term of the even numbers is greater than 101 and lesser than 200, how many such sequences can be formed?

(A) 12
(B) 17
(C) 25
(D) 33
(E) 50

Originally posted by bindiyajoisher on 21 Sep 2011, 11:52.
Last edited by Bunuel on 25 Sep 2013, 01:36, edited 1 time in total.
RENAMED THE TOPIC.
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 10011
Location: Pune, India

### Show Tags

22 Sep 2011, 03:34
69
58
bindiyajoisher wrote:
The sum of four consecutive odd numbers is equal to the sum of 3 consecutive even numbers. Given that the middle term of the even numbers is greater than 101 and lesser than 200, how many such sequences can be formed?
(A) 12
(B) 17
(C) 25
(D) 33
(E) 50

Sum of four consecutive odd numbers:
(2a - 3) + (2a - 1) + (2a + 1) + (2a + 3) = 8a

Sum of three consecutive even numbers:
(2b - 2) + 2b + (2b + 2) = 6b

Given 8a = 6b or a/b = 3/4, a and b can be any integers. So, 'a' has to be a multiple of 3 and 'b' has to be a multiple of 4. Possible solutions are: a = 3, b = 4; a = 6, b = 8; a = 9, b = 12 etc
Since 101 < 2b < 200 i.e.
51 <= b < 100
Since b also has to be a multiple of 4, the values that b can take are 52, 56, 60, 64 ... 96
Number of values b can take = (Last term - First term)/Common Difference + 1 = (96 - 52)/4 + 1 = 12
_________________
Karishma
Veritas Prep GMAT Instructor

Intern
Status: Need to read faster and get less distracted by the ticking clock!
Affiliations: Golden Key
Joined: 19 Nov 2010
Posts: 11
Location: Toronto, ON

### Show Tags

21 Sep 2011, 13:27
28
3
12
A

1+3+5+7 = 16
3+5+7+9 = 24
5+7+9+11=32 ... so you can see the sum of 4 odd numbers increment by 8.

Therefore all even number that are divisible by 8 between 101-200 will have a possible series that adds up to it.. therefore 200-101/8 gives you 12.75 = 12!
##### General Discussion
Retired Moderator
Joined: 20 Dec 2010
Posts: 1535

### Show Tags

21 Sep 2011, 12:46
12
11
bindiyajoisher wrote:
The sum of four consecutive odd numbers is equal to the sum of 3 consecutive even numbers. Given that the middle term of the even numbers is greater than 101 and lesser than 200, how many such sequences can be formed?
(A) 12
(B) 17
(C) 25
(D) 33
(E) 50

Four consecutive odd numbers: k-2, k, k+2, k+4
Three consecutive even numbers: n-2, n, n+2

k-2+k+k+2+k+4=n-2+n+n+2
4k+4=3n
4(k+1)=3n
k+1=(3/4)n
k=(3/4)n-1

All n's that's divisible by 4 will have an integral k. So, we need to find out how many such n's are available within given range:

We know,
101<n<200
104<=n<=196

Count=(196-104)/4+1=92/4+1=23+1=24

Ans: 24.
Retired Moderator
Joined: 20 Dec 2010
Posts: 1535

### Show Tags

21 Sep 2011, 12:55
bindiyajoisher wrote:
but the OA is 12

Yeah!!! So I see. I'll wait for few others to point out the error. I can't think where I may be going wrong. BTW, what's the source of the problem?
Intern
Joined: 09 Aug 2011
Posts: 7
Location: India
Concentration: Finance, Strategy
GMAT Date: 09-30-2011
WE: Corporate Finance (Energy and Utilities)

### Show Tags

21 Sep 2011, 13:01
it is from the test from the institute I have enrolled
Intern
Joined: 01 Nov 2010
Posts: 7
Location: United States
GMAT Date: 03-30-2012
WE: Pharmaceuticals (Health Care)

### Show Tags

21 Sep 2011, 13:17
1
2
fluke wrote:
bindiyajoisher wrote:
The sum of four consecutive odd numbers is equal to the sum of 3 consecutive even numbers. Given that the middle term of the even numbers is greater than 101 and lesser than 200, how many such sequences can be formed?
(A) 12
(B) 17
(C) 25
(D) 33
(E) 50

Four consecutive odd numbers: k-2, k, k+2, k+4
Three consecutive even numbers: n-2, n, n+2

k-2+k+k+2+k+4=n-2+n+n+2
4k+4=3n
4(k+1)=3n
k+1=(3/4)n
k=(3/4)n-1

All n's that's divisible by 4 will have an integral k. So, we need to find out how many such n's are available within given range:

We know,
101<n<200
104<=n<=196

Count=(196-104)/4+1=92/4+1=23+1=24

Ans: 24.

k-2, k, k+2, k+4 n-2, n, n+2

Why did you take the minus??? Why not :
Four consecutive odd numbers: O, O+2, O+4, O+6
Three consecutive even numbers: E, E+2, E+4
O+O+2+ O+4+ O+6 = E+ E+2+E+4
4O + 12 = 3E + 6
4O + 6 = 3E
4O = 3(E-2)
O = 3(E-2)/4
Retired Moderator
Joined: 20 Dec 2010
Posts: 1535

### Show Tags

21 Sep 2011, 13:41
5
3
fluke wrote:
bindiyajoisher wrote:
The sum of four consecutive odd numbers is equal to the sum of 3 consecutive even numbers. Given that the middle term of the even numbers is greater than 101 and lesser than 200, how many such sequences can be formed?
(A) 12
(B) 17
(C) 25
(D) 33
(E) 50

Four consecutive odd numbers: k-2, k, k+2, k+4
Three consecutive even numbers: n-2, n, n+2

k-2+k+k+2+k+4=n-2+n+n+2
4k+4=3n
4(k+1)=3n
k+1=(3/4)n
k=(3/4)n-1

All n's that's divisible by 4 will have an integral k. So, we need to find out how many such n's are available within given range:

We know,
101<n<200
104<=n<=196

Count=(196-104)/4+1=92/4+1=23+1=24

Ans: 24.

Yes, now I understand.

(3/4)n should be even to make the k an ODD integer; I just considered integer before.

Thus, n must have at least three 2's in it, or n must be divisible by 8.

Count of numbers divisible by 8 between 101 and 200 is 12.

Ans: "A"
**********************

thanks folks
Manager
Status: Still Struggling
Joined: 03 Nov 2010
Posts: 108
Location: India
GMAT Date: 10-15-2011
GPA: 3.71
WE: Information Technology (Computer Software)

### Show Tags

22 Sep 2011, 00:34
1
Hi,
pls help me proceed, my style...or tell me wats wrong in my approach:

odd numbers : (2a+1)+(2a+3)+(2a+5)+(2a+7) = 8a +16
even numbers:(2b)+(2b+2)+(2b+4) = 6b + 6
hence,
8a+16=6b+6
a= (3b-5)/4

..how to proceed from here?
Manager
Status: Essaying
Joined: 27 May 2010
Posts: 79
Location: Ghana
Concentration: Finance, Finance
Schools: Cambridge
GMAT 1: 690 Q47 V37
GPA: 3.9
WE: Accounting (Education)

### Show Tags

22 Sep 2011, 02:03
fluke can you elaborate...
Retired Moderator
Joined: 20 Dec 2010
Posts: 1535

### Show Tags

22 Sep 2011, 03:08
4
4
liftoff wrote:
fluke can you elaborate...

The idea is simple:

If k is odd
AND n is even.

Why did I consider:
n-2, n, n+2: For the sake of simplicity, because we are given that
101<n<200 (Middle term of the even number)

According to given condition:
Sum of 4 consecutive integers=Sum of 3 even integers
k-2+k+k+2+k+4=n-2+n+n+2
4k+4=3n
4(k+1)=3n
k=(3/4)n-1;

Now, we know that n(middle term of the even sequence) is between 101 and 200, exclusive

So, 101<n<200

But, we should also conform to the fact that k is ODD.

How can we get k as odd
Say n=102;
k=(3/4)*102-1=76.5-1=75.5(It is NOT ODD); so n=102 IS not a possible/valid sequence
Likewise n=103; will also not give k as odd;
n=104;
k=(3/4)*104-1=77(ODD)

We see that if n=A multiple of 8, then k becomes ODD. How so?

4*Even=4*2x=Even
So, n must be in the form of 8x.

105,106,107,108,109,110,111(They are not divisible by 8), thus k can't be an ODD integer

112 is divisible by 8.

So, if we find all the values from 101 to 200 that are divisible by 8, we will have our count. The sequence will be.

1st: 102,104,106
2nd: 110,112,114
3rd: 118,120,122
...
12th: 190,192,194
Note: we just have to care about the middle term. The first term and last term will be follow: +-2.

Also, so far k is ANY ODD integer, we are good.
Intern
Joined: 14 Oct 2012
Posts: 5
Re: The sum of four consecutive odd numbers is equal to the sum  [#permalink]

### Show Tags

13 Nov 2013, 02:48
3
1
bindiyajoisher wrote:
The sum of four consecutive odd numbers is equal to the sum of 3 consecutive even numbers. Given that the middle term of the even numbers is greater than 101 and lesser than 200, how many such sequences can be formed?

(A) 12
(B) 17
(C) 25
(D) 33
(E) 50

We need to choose n : 101 < n < 200 and n is even, and (n-2) + n + (n+2) = sum of 4 consecutive odd numbers.

I am assuming the 4 consecutive odd numbers to be e-3, e-1, e+1, e+3 where e could be an even integer > 3.

Hence: (n-2) + n + (n+2) = e-3 + e-1 + e+1 + e+3

3n = 4e

Let us analyse the above equation. For the above equation to hold true, e should be a multiple of 6 (as it should have a 3 in it and is an even) and n should be a multiple of 8 (because n should have factors 4 and 2 as 3 is a prime already and 4e has these factors in RHS). This is the least requirement.

So every multiple of 8 will satisfy the above equation if I do not restrict e.

So possible values of n(for no restriction on e) = all multiples of 8 between 101 and 200 = 12.

Hope I made my point clear
Senior Manager
Joined: 07 Aug 2011
Posts: 494
GMAT 1: 630 Q49 V27
The sum of four consecutive odd numbers is equal to the sum  [#permalink]

### Show Tags

12 Apr 2015, 09:37
1
krishnasty wrote:
Hi,
pls help me proceed, my style...or tell me wats wrong in my approach:

odd numbers : (2a+1)+(2a+3)+(2a+5)+(2a+7) = 8a +16
even numbers:(2b)+(2b+2)+(2b+4) = 6b + 6
hence,
8a+16=6b+6
a= (3b-5)/4

..how to proceed from here?

odd numbers : (2a+1)+(2a+3)+(2a+5)+(2a+7) = 8a +16 = $$8K_1$$
even numbers:(2b)+(2b+2)+(2b+4) = 6b + 6= 6(b+1)

we have to find number which is both multiple of 6 and 8.

lets looks at the minimum sum
102+104+106 = 312 = 75+77+79+81

lets looks at maximum sum = 198+200 +202 = 600=197+ 199 + 201 + 203

so 312<=6(b+1) <= 600 and 6(b+1) should be divisible by 6 and 8 .
51<=b<=99, off these values whenever b+1 is multiple of 4 , 6(b+1) will be divisible by 8 and 6 too .

52 is first term and 96 is last , N= 12 .
hope it helps .
Manager
Joined: 09 Jun 2015
Posts: 78
Re: The sum of four consecutive odd numbers is equal to the sum  [#permalink]

### Show Tags

17 Apr 2016, 08:04
1
bindiyajoisher wrote:
The sum of four consecutive odd numbers is equal to the sum of 3 consecutive even numbers. Given that the middle term of the even numbers is greater than 101 and lesser than 200, how many such sequences can be formed?

(A) 12
(B) 17
(C) 25
(D) 33
(E) 50

2x-1+2x+1+2x+3+2x+5=2y+2y+2+2y+4
8x+8=6y+6
6y=8x+2
2y=(8x+2)/3
when x=2, y=3

The first set is 3+5+7+9=6+8+10
The second set is 9+11+13+15 = 14+16+18
Third set is 15+17+19+21 = 22+24+26
If you observe carefully, you get the pattern. Right hand side is what we want. The even numbers start with 6 and the next numbers 8 added to the previous number.
We get 12 numbers between 1 and 100. Similarly, we get 12 numbers between 101 and 200.
VP
Joined: 07 Dec 2014
Posts: 1229
The sum of four consecutive odd numbers is equal to the sum  [#permalink]

### Show Tags

Updated on: 06 Aug 2018, 11:37
The sum of four consecutive odd numbers is equal to the sum of 3 consecutive even numbers. Given that the middle term of the even numbers is greater than 101 and lesser than 200, how many such sequences can be formed?

(A) 12
(B) 17
(C) 25
(D) 33
(E) 50

least two such sequence pairs are:
3, 5, 7, 9 and 6, 8, 10, both summing to 24; and
9, 11, 13, 15 and 14, 16, 18, both summing to 48
middle term of even sequence is a multiple of 8
least multiple of 8>101=104
greatest multiple of 8<200=192
let x=number of such sequences
104+8(x-1)=192
8x=96
x=12
A

Originally posted by gracie on 17 Apr 2016, 14:13.
Last edited by gracie on 06 Aug 2018, 11:37, edited 2 times in total.
Intern
Joined: 26 Sep 2015
Posts: 4
Re: The sum of four consecutive odd numbers is equal to the sum  [#permalink]

### Show Tags

29 May 2016, 15:59
5
3
My ELI5 version:

Find a pattern.
1+3+5+7 = 16
3+5+7+9 = 24
5+7+9+11 = 32

From these we have determined that the sums will always be limited to a multiple of 8.

Now we take a look at the limitation set by the problem.
101<n<200

The first multiple of 8 that is greater than 101 is 104.
The last multiple of 8 that is less than 200 is 192.
192-104 = 88
88/8 = 11
11+1(adding back the 104) = 12
Intern
Joined: 20 Oct 2015
Posts: 36
Re: The sum of four consecutive odd numbers is equal to the sum  [#permalink]

### Show Tags

06 Jul 2016, 03:30
VeritasPrepKarishma wrote:
bindiyajoisher wrote:
The sum of four consecutive odd numbers is equal to the sum of 3 consecutive even numbers. Given that the middle term of the even numbers is greater than 101 and lesser than 200, how many such sequences can be formed?
(A) 12
(B) 17
(C) 25
(D) 33
(E) 50

Sum of four consecutive odd numbers:
(2a - 3) + (2a - 1) + (2a + 1) + (2a + 3) = 8a

Sum of three consecutive even numbers:
(2b - 2) + 2b + (2b + 2) = 6b

Given 8a = 6b or a/b = 3/4, a and b can be any integers. So, 'a' has to be a multiple of 3 and 'b' has to be a multiple of 4. Possible solutions are: a = 3, b = 4; a = 6, b = 8; a = 9, b = 12 etc
Since 101 < 2b < 200 i.e.
51 <= b < 100
Since b also has to be a multiple of 4, the values that b can take are 52, 56, 60, 64 ... 96
Number of values b can take = (Last term - First term)/Common Difference + 1 = (96 - 52)/4 + 1 = 12

awesome explanation thanks million Karishma

yet I have a qustion
since 101<2b<100
why b must be equal to 51? because 2b must be at least 102?thats why?

can we say 52<= b<= 98 ? isnit better?

thanks again.
Kamran
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 10011
Location: Pune, India
Re: The sum of four consecutive odd numbers is equal to the sum  [#permalink]

### Show Tags

06 Jul 2016, 21:01
1
kamranko wrote:
VeritasPrepKarishma wrote:
bindiyajoisher wrote:
The sum of four consecutive odd numbers is equal to the sum of 3 consecutive even numbers. Given that the middle term of the even numbers is greater than 101 and lesser than 200, how many such sequences can be formed?
(A) 12
(B) 17
(C) 25
(D) 33
(E) 50

Sum of four consecutive odd numbers:
(2a - 3) + (2a - 1) + (2a + 1) + (2a + 3) = 8a

Sum of three consecutive even numbers:
(2b - 2) + 2b + (2b + 2) = 6b

Given 8a = 6b or a/b = 3/4, a and b can be any integers. So, 'a' has to be a multiple of 3 and 'b' has to be a multiple of 4. Possible solutions are: a = 3, b = 4; a = 6, b = 8; a = 9, b = 12 etc
Since 101 < 2b < 200 i.e.
51 <= b < 100
Since b also has to be a multiple of 4, the values that b can take are 52, 56, 60, 64 ... 96
Number of values b can take = (Last term - First term)/Common Difference + 1 = (96 - 52)/4 + 1 = 12

awesome explanation thanks million Karishma

yet I have a qustion
since 101<2b<100
why b must be equal to 51? because 2b must be at least 102?thats why?

can we say 52<= b<= 98 ? isnit better?

thanks again.
Kamran

Since 2b must be greater than 101, 2b must be at least 102 i.e. b must be at least 51.
Since 2b must be less than 200, b must be less than 100 so b can be 99 at the most.
_________________
Karishma
Veritas Prep GMAT Instructor

Intern
Joined: 13 Apr 2016
Posts: 2
Concentration: Sustainability, Technology
Schools: LBS '18
GMAT 1: 740 Q48 V44
WE: Business Development (Energy and Utilities)
Re: The sum of four consecutive odd numbers is equal to the sum  [#permalink]

### Show Tags

15 Sep 2016, 08:58
hashjax wrote:
A

1+3+5+7 = 16
3+5+7+9 = 24
5+7+9+11=32 ... so you can see the sum of 4 odd numbers increment by 8.

Therefore all even number that are divisible by 8 between 101-200 will have a possible series that adds up to it.. therefore 200-101/8 gives you 12.75 = 12!

Hey guys,

I understand everything about all your solutions, I just don't understand how 200-101/8 gives you the numbers divisible by 8 between 101-200..

Can you explain this?

Thanks,

Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 10011
Location: Pune, India
Re: The sum of four consecutive odd numbers is equal to the sum  [#permalink]

### Show Tags

15 Sep 2016, 21:36
2
damafisch wrote:
hashjax wrote:
A

1+3+5+7 = 16
3+5+7+9 = 24
5+7+9+11=32 ... so you can see the sum of 4 odd numbers increment by 8.

Therefore all even number that are divisible by 8 between 101-200 will have a possible series that adds up to it.. therefore 200-101/8 gives you 12.75 = 12!

Hey guys,

I understand everything about all your solutions, I just don't understand how 200-101/8 gives you the numbers divisible by 8 between 101-200..

Can you explain this?

Thanks,

It doesn't and you shouldn't use this method.
Multiples of 8 lying between two integers should be calculated as below:
The first multiple of 8 in the range: 8 * 13 = 104
The last multiple of 8 in the range: 8*24 = 192

So all multiples from the 13th to the 24th multiple of 8 are in the range. Number of multiples = 24 - 13 + 1 = 12
_________________
Karishma
Veritas Prep GMAT Instructor

Re: The sum of four consecutive odd numbers is equal to the sum   [#permalink] 15 Sep 2016, 21:36

Go to page    1   2    Next  [ 36 posts ]

Display posts from previous: Sort by