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.