PS -

If my reasoning is right, you should find the logic of the sequence :
a1=64
a2=66
a3=67
a4=64+8=72
A5=66+8=74
a6=67+8=75
A7=72+8=80
a8=74+8=82
a9=75+8=83
a10=80+8=88
a11=82+8=90
a12=83+8=91
look , it is +2+1+5, +2+1+5,+2+1+5.
soo, if i am correct we are able to find any number of a sequence using following formula:
Tn=64+(2+1+5)K, Tn=64+8k - if the number is composed by the whole sequence of +2+1+5. So after a subtractment of 64 and devision by 8, there should be an integer, or an integer + remainder of 2 or 3.
Lets backsolve:
E:remainder 7 - can't be
D:remainder 1
C:remainder 6
B:remainder 5
A:remainder 2, i think A is the right answer.

Oh my. Just figured it out. Forget all my previous solution.
It is very similar to the sequence where we have only a1, that differs from other terms of sequence. i.e. all remaining will be a2=a1+x,a3=a2+x and so on.
So, knowing that sequence is +2+1+5=+8.
Then, you need just to backsolve from he answers.
Just subtract 64,66 and 67 from the answer choice and then divide it by 8. Which (answer-64-66-67)will be divisible by 8 perfectly, it ll be the right one.

a) 762-64-66-67=565, 565/8 = not integer, so wrong
b) 765-64-66-67=568, 568/8=71, GOTCHA !
Answer is B, not A. sorry.

OA is A
Could you find a more simple way to resolve it ?

Xbond wrote:Hi

how do resolve thsi PS in fast way ?

Hi xbond,

please post problem solving questions in the problem solving forum.

In these kind of problems, applying some pattern analysis will yield dividends...don't go to the answer choices prematurely!

We know that any term is 8 greater than the one that came third before it.

Run the sequence once to find the pattern:
a1 = 64; a2 = 66; a3 = 67; a4 = 72

So, a number 'n' is a term in the sequence if n is a multiple of 8, if n-2 is a multiple of 8, if n-3 is a multiple of 8, or if n+5 is a multiple of 8.

Let's look at the choices:
choice A: 762

762 is not a multiple of 8. But 762-2 = 760 is. So this is a term in the sequence, and we are done.

Choose A

(and, of course, if n+5 is a multiple of 8, then so is n-3)

IMO A

the series is
64,66,67,72,74,75,80,82,83......so.on

if we look closely then if n is a term and if it is even then either it should be div by 8 or n-2 shud be div by 8

but if it is odd then it n-3 should be div by 8

now check the options
a)762 even not div by 8 ; but 762-2=760 div by 8
b)765 odd 765-3 not div by 8
c)801 odd 801-3 not div by 8
d)1006 even not div by 8 and 1004 not div by 8
e)1287 odd 1287-3 not div by 8

hence 762 is the ans as it satisfy the condition

xcusemeplz2009 wrote:IMO A

the series is
64,66,67,72,74,75,80,82,83......so.on

if we look closely then if n is a term and if it is even then either it should be div by 8 or n-2 shud be div by 8

but if it is odd then it n-3 should be div by 8

now check the options
a)762 even not div by 8 ; but 762-2=760 div by 8
b)765 odd 765-3 not div by 8
c)801 odd 801-3 not div by 8
d)1006 even not div by 8 and 1004 not div by 8
e)1287 odd 1287-3 not div by 8

hence 762 is the ans as it satisfy the condition

That is an excellent approach xcusemeplz2009!

But remember, once you found that choice A worked, you were done: no need to check the rest of the choices because there cannot be two correct answers.