Hi, t1 is odd t2 is odd t3 is even t4 = t3+t2 = even+odd = odd t5 = t4+t3 = odd+even = odd t6 = t5+t4 = odd+odd = even t7 = t6+t5 = even+odd = odd t8 = t7+t6 = odd+even = odd t9 = t8+t7 = odd+odd = even . . You can see the pattern, every third element is even and others are odd. So, if there are 300...

bblast wrote: Quick question - Do we still call the set-"an evenly spaced set" after removing the other pairs (other than 1,19) ? I hope NO ?

I didn't understand statement 2. Does it mean that the sum of the n consecutive integers is a multiple of 9? Yes, but not directly. If the n consecutive integers are 2,3,4,5,6,7.. then 234567(6- digit number) is divisible by 9. Anyway for a number to be divisible by 9, the sum of its digits should ...

Condition 2 -- Consider 11+12+13 = 36 -even Consider 2+3+4+5+6+7 = 27 -- ODD... Not sufficient. Hence answer = A Hi, Minor mistake. You cannot choose 11,12,13. If you are choosing this set, you are picking n as 3. When we write those numbers in a series it becomes 111213, which is not 'n' digit num...

From (1) A(A²-1)<0 Meaning either A>0 and (A²-1)<0 Or A<0 and (A²-1)>0 This gives us conflicting values of A and hence INSUFFICIENT. Is my reasoning sufficient? Do I need to do any further testing? Hi, I don't clearly understand what you mean by conflicting values. Anyway, my take on this: a>0 a...

Hi,
From(1):
Sum of integers from a to (a+5) = 6a+15. This is always odd.
Sufficient

From(2):
if the number is 45, sum is odd
if the number is 3456, sum is even
Not sufficient

Hence, A

Hi, Bag A -> Red : White = 1:3. So, white is a multiple of 3 White : Blue = 2:3. So, white is a multiple of 2 So, white is a multiple of 2 and 3. So, Red:White:Blue = 2:6:3 or 2a:6a:3a Bag B -> Red : White = 1:4 or b:4b 6a+4b = 30 So, only possible positive integral solutions are: a=1,b=6 a=3,b=3 So...

Totally agree mate. That is the trickiest part indeed, at least for me. Practice and alertness should make us better!