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

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

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

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?

To elaborate more.

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.

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

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 .

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.

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

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.

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

VeritasPrepKarishma pls verify this method, you may be able to put it more clearly I guess
here's my attempt:

SUM = average * number of elements
sum depends on average , lets find the average
average of 4 consecutive odd integers must be even
for eg: 75,77,79,81 here the Avg is the middle no. 78
{Basically, 4 consecutive positive odd integers can be formed if we have an even no.>4 as the average.}
With this understanding ,
We have middle no. of 3 consecutive even no.s(i.e. average) ranging from 102 to 198 (inc.)
How many even no.s can be formed using above no.s , lets see
102/4 - not integer
104/4 = 26 ( even)
106/4 - not integer
108/4 - 27 ( divisible, but not even!)
110/4 - not integer
112/4 - 28 (even)..........
If you see the pattern here we get the number we want at the gap of 8 starting from 104
so 104,112,120,.......192
{take for eg: With 120 we can form 4 consecutive odd integers by using 120 as the average - 117,119,121,123}

count the no. of middle elements = 192-104/8 +1 = 12

Yes Deepak, it's perfectly fine.
The way I see it, you are saying that the mean of 4 odd integers will be even. This mean will be equal to 3*Mean of even integers/4.
Since 3 is not divisible by 4, Mean of even integers should be divisible by 4.
So for every such value of mean, you will have a corresponding sequence of 4 odd integers.

VeritasPrepKarishma

Thanks Karishma. Your posts always help and I very well understood the solution after reading your post. As I started solving the question in the above way(that means, I took even and odd numbers as taken by Lucky), my mind keeps on thinking how can this be solved if we take these number.

I tried to understand his solution. Please help why did he take the smallest middle number 104(as highlighted in red), the smallest middle number could be 102. Please correct me if I am wrong. I could not think of converting the sum of odd numbers to \(8K_1\). Please tell how can this be solved when the below even and odd numbers are considered?

odd numbers : (2a+1)+(2a+3)+(2a+5)+(2a+7)
even numbers:(2b)+(2b+2)+(2b+4)

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

\(8a+16=6b+6\)
\(b + 1 = \frac{4*(a+2)}{3}\)

So (b + 1) is a multiple of 4.

"Given that the middle term of the even numbers is greater than 101 and lesser than 200" - Middle term of even numbers = 2b + 2
The minimum value it can take is 102 so minimum value of b + 1 is 51.
But, b+1 needs to be a multiple of 4. Hence, the minimum value of b+1 would be 52 and hence the minimum value of the minimum term would be 104.
Hope this clarifies.