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

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

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21 Sep 2011, 11:52
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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
[Reveal] Spoiler: OA

Last edited by Bunuel on 25 Sep 2013, 01:36, edited 1 time in total.
RENAMED THE TOPIC.
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21 Sep 2011, 12:46
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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.
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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?
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21 Sep 2011, 13:01
it is from the test from the institute I have enrolled
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21 Sep 2011, 13:17
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
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21 Sep 2011, 13:27
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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!
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21 Sep 2011, 13:41
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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
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22 Sep 2011, 00:34
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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?
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22 Sep 2011, 02:03
fluke can you elaborate...
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22 Sep 2011, 03:08
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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.
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22 Sep 2011, 03:34
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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
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24 Sep 2013, 15:47
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Re: The sum of four consecutive odd numbers is equal to the sum [#permalink]

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13 Nov 2013, 02:48
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
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Re: The sum of four consecutive odd numbers is equal to the sum [#permalink]

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12 Apr 2015, 04:03
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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

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12 Apr 2015, 09:37
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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 .
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Re: The sum of four consecutive odd numbers is equal to the sum [#permalink]

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17 Apr 2016, 08:04
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.
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The sum of four consecutive odd numbers is equal to the sum [#permalink]

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17 Apr 2016, 14:13
the sums must be a multiple of 8=(4)(2) and 6=(3)(2)
brief trial and error confirms that sums are equal to multiples of 24
the ratio of even sum to middle even term=3:1
even sums must be＞(3)(101)=303 and＜(3)(200)=600
only 12 multiples of 24 occur within this range
12 sequences can be formed
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Re: The sum of four consecutive odd numbers is equal to the sum [#permalink]

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29 May 2016, 15:59
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
Re: The sum of four consecutive odd numbers is equal to the sum   [#permalink] 29 May 2016, 15:59
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