@rachit_28 said:24, 60, 120, 210, ?300336420525
336?
@chillfactor said:
P is a set of n consecutive integers such that sum of digits of any of these n numbers is not divisible by 8. Find the maximum possible value of n.
14 ?
9999993 to 10000006
@Cat.Aspirant123 said:Bacteria reproduce in such a manner that every hour their number doubles. In a controlledexperiment that started with one bacterium in a jar at 2 pm on Saturday, the jar was found to becompletely full at 12 noon on Wednesday. Had there been 64 bacteria initially, when would the jar becompletely full?
6am wed?
@Cat.Aspirant123 said:Bacteria reproduce in such a manner that every hour their number doubles. In a controlledexperiment that started with one bacterium in a jar at 2 pm on Saturday, the jar was found to becompletely full at 12 noon on Wednesday. Had there been 64 bacteria initially, when would the jar becompletely full?
2^0 pe 12 noon
2^6 pe 12 - 6 = 6 am
@arumugadas said:there are 10 pockets of chips and biscuits in the cupboard. ram pulls out 3 pockets and all of them turn out to be chips what is the probability that the snack cupboard contain 1 packet of biscuits and 9 pockets of chips
@Torque024 said:How you got this?
I started with 1 to 7 .. but my sixth sense said that 7 is too small a answer
I moved to a higher number ,100 and I noticed that there is drastic change in the sum of digits from 99 to 100, and that 98 to 106 are consecutive and each sum of digits is not divisible by 8
So, I moved on to higher order, 997 to 1006 and 9996 to 1006 and so on...
But if I make the starting number i.e. the 9999... number an eight digit number, I lose the advantage as 9 written 8 times is 72 and it breaks the consecutive sequence, so I restricted it to seven digit and that's when I started with 9 written 6 times followed by 3 , because 9999992 's sum of digits is divisible by 8
So, I moved from 9999993 to 10000006 , hence 14 numbers
If you move further up beyond 9 digit numbers, the series repeats I think
If S = 3 + 5 + 9 + 17 + 33 + €Ś the value of the 10th term of the series is _____.
@arumugadas said:
10 chips, 0 biscuits = 120/120 In any case we have to pick 3 chips.
9 chips, 1 biscuit 9C3/10C3 = 84/120
8 chips, 2 biscuit 8C3/10C3 = 56/120
P = 84/(120+84+56+35+20+10+4+1) = 14/55
Was done previously.
9 chips, 1 biscuit 9C3/10C3 = 84/120
8 chips, 2 biscuit 8C3/10C3 = 56/120
P = 84/(120+84+56+35+20+10+4+1) = 14/55
Was done previously.
@Cat.Aspirant123 said:If S = 3 + 5 + 9 + 17 + 33 + €Ś the value of the 10th term of the series is _____.
1025?
@Cat.Aspirant123 said:If S = 3 + 5 + 9 + 17 + 33 + €Ś the value of the 10th term of the series is _____.
1025
@Cat.Aspirant123 said:If S = 3 + 5 + 9 + 17 + 33 + €Ś the value of the 10th term of the series is _____.
1025..
@Cat.Aspirant123 said:If S = 3 + 5 + 9 + 17 + 33 + €Ś the value of the 10th term of the series is _____.
1025
@Torque024 said:How you got this?
Sum of digits of a number n = k
Then sum of digits of (n + 1) will be:-
k + 1, if unit digit of n is not 9
and k + 1 - 9a, where a is the number 9's at the end of n
So, clearly we can have atmost 7 + 7 numbers such that sum of digits of none of these numbers is divisible by 8
Try the same question for sum of digits divisible by 10 or 11 (if you are interested, it won't appear in any of the MBA entrance examination)
@Cat.Aspirant123 said:how
3,5,9,17,33,65,129,257,513,1025
Observe every term its equal to 2t-1 where t is preceding term