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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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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, 10:52.
Last edited by Bunuel on 25 Sep 2013, 00:36, edited 1 time in total.
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Re: Consecutive numbers  [#permalink]

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New post 22 Sep 2011, 02:34
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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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Re: Consecutive numbers  [#permalink]

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New post 21 Sep 2011, 12:27
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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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Re: Consecutive numbers  [#permalink]

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New post 21 Sep 2011, 11:46
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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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Re: Consecutive numbers  [#permalink]

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New post 21 Sep 2011, 11:55
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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Re: Consecutive numbers  [#permalink]

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New post 21 Sep 2011, 12:01
it is from the test from the institute I have enrolled
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Re: Consecutive numbers  [#permalink]

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New post 21 Sep 2011, 12:17
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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.



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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Re: Consecutive numbers  [#permalink]

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New post 21 Sep 2011, 12:41
5
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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"
**********************

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Re: Consecutive numbers  [#permalink]

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New post 21 Sep 2011, 23:34
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?
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Re: Consecutive numbers  [#permalink]

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New post 22 Sep 2011, 01:03
fluke can you elaborate...
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Re: Consecutive numbers  [#permalink]

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New post 22 Sep 2011, 02:08
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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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Re: The sum of four consecutive odd numbers is equal to the sum  [#permalink]

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New post 13 Nov 2013, 01:48
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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


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

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New post 12 Apr 2015, 08:37
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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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New post 17 Apr 2016, 07:04
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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

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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New post Updated on: 06 Aug 2018, 10:37
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, 13:13.
Last edited by gracie on 06 Aug 2018, 10:37, edited 2 times in total.
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Re: The sum of four consecutive odd numbers is equal to the sum  [#permalink]

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New post 29 May 2016, 14:59
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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
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Re: The sum of four consecutive odd numbers is equal to the sum  [#permalink]

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New post 06 Jul 2016, 02:30
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?that`s why?

can we say 52<= b<= 98 ? is`nit better?

thanks again.
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Re: The sum of four consecutive odd numbers is equal to the sum  [#permalink]

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New post 06 Jul 2016, 20:01
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?that`s why?

can we say 52<= b<= 98 ? is`nit 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.
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Re: The sum of four consecutive odd numbers is equal to the sum  [#permalink]

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New post 15 Sep 2016, 07:58
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,

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

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New post 15 Sep 2016, 20:36
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,

Adam


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
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