Re: In a sequence of 700 integers, each term after the first two terms is
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22 Sep 2022, 07:02
Say the first number in the sequence is a1 and the second a2. We can think of these two in terms of Odd and Even number and list the choices.
Case I
a1 and a2 can both be even.
Since subsequent numbers are found by additions, all numbers are even. Therefore, there will no odd numbers. Choice A.
Case II
a1 can be odd and a2 even.
This leads to a3 to be odd, a4 odd since
a3 = a1 + a2
Odd + Even = Odd
a4 = Odd + Even = Odd
Continuing this way,
a5 = Even
a6 = Odd
a7 = Odd
a8 = Even
..
Notice the pattern is: (Odd, Even, Odd), (Odd, Even, Odd), (Odd, Even, Odd)
The pattern is a repeating sequence of Odd, Even, Odd.
The number of this sequence is 700/3 = 233 1/3. Which is 699 + 1.
First 699 numbers will have 2/3 Odds. This equals: 699*2/3 = 466. The 700th number is also an Odd.
Therefore, the total is 467. Choice G.
Case III
a1 can be even and a2 odd.
The pattern now is :
a1 Even,
a2 Odd,
a3 Odd,
a4 Even,
a5 Odd
a6 Odd
a7 Even
Notice the the pattern (Even, Odd, Odd) repeats 233 times. And there remains the 700th number. The 700th number now is an Even number.
Therefore, total number of odds now is 466. Choice F.
Case IV
Both a1 and a2 are Odd.
The sequence is:
a1 Odd
a2 Odd
a3 Even
a4 Odd
a5 Odd
a6 Even
The sequence has like before 466 odd numbers. The 700th number is an odd number. So, total number of odds is 467. Choice G.