@Logrhythm said:bhai 5^anything = xxxx25where anything > 1
haan yaar....pehele dimaag nahi chala...
@hatemonger said:Here no of teenagers are not given ok but if total no of persons are given then can we do it by finding min value of summation of ages... when product is constant summation is min when numbers are equal..... right approach or not ???
Product has to satisfy....regards
scrabbler
@aditi88 said:the product of ages of some teenagers is 10584000. the sum of their ages??A>86B>88C>89D>87plz quote me..
10584000 = 2^3*5^3*2^3*3^3*7^2 = 2^6*3^3*5^3*7^2
we have six 2's, three 3's, three 5's and two 7's
7 cannot go with either of 5 or 3
so two of the ages are 7*2 and 7*2
left with four 2's, three 3's and three 5's
5*2 is not valid and neither is 5*4 or any other power of 2 with 5
so three of the other ages are 5*3, 5*3 and 5*3
and one other age is 2^4
so sum = 16+(15*3)+(14*2) = 89....
If a number is chosen at random from integer 1 to 96 inclusive wat is the probability that n (n +1) (n+ 2) will be divisible by 8?
a. 1/4
b. 3/8
c. 1/2
d. 5/8
e. 3/4
a. 1/4
b. 3/8
c. 1/2
d. 5/8
e. 3/4
@hedonistajay said:If a number is chosen at random from integer 1 to 96 inclusive wat is the probability that n (n +1) (n+ 2) will be divisible by 8?a. 1/4b. 3/8c. 1/2d. 5/8e. 3/4
60/96 = 5/8...
@hedonistajay said:If a number is chosen at random from integer 1 to 96 inclusive wat is the probability that n (n +1) (n+ 2) will be divisible by 8?a. 1/4b. 3/8c. 1/2d. 5/8????e. 3/4
@hedonistajay said:If a number is chosen at random from integer 1 to 96 inclusive wat is the probability that n (n +1) (n+ 2) will be divisible by 8?a. 1/4b. 3/8c. 1/2d. 5/8e. 3/4
5/8
@aditi88 said:the product of ages of some teenagers is 10584000. the sum of their ages??A>86B>88C>89D>87plz quote me..
89 ?
10584000 = 2^6*3^3*5^3*7^2
3 teenagers of age = 15
2 of age = 14
1 of age = 16
3*15 + 2*14 + 16 = 89
@hedonistajay said:If a number is chosen at random from integer 1 to 96 inclusive wat is the probability that n (n +1) (n+ 2) will be divisible by 8?a. 1/4b. 3/8c. 1/2d. 5/8e. 3/4
Case 1: n is even = 48 cases
Case 2: n is odd = 8k-1 = 12 numbers
Pro = 60/96 = 5/8
every sequence starting wid even number will satisfy this... (2 3 4)(4 5 6)............. there are 96/2=48
also every triplet that has multiple of 8 will satisfy..(7 8 9)(15 16 17) .......... there are 96/8=12
total =60
60/96
@hedonistajay said:how 5/8 .. kindly explain
n should be a multiple of 2 to make n+2 a multiple of 4
so 48 such numbers are there
plus n+1 can be a multiple of 8, so 12 such numbers...
total = 60
Q1
Q 2
@hedonistajay said:If a number is chosen at random from integer 1 to 96 inclusive wat is the probability that n (n +1) (n+ 2) will be divisible by 8?a. 1/4b. 3/8c. 1/2d. 5/8e. 3/4
@catahead said:It is 5/8Just saw it QEhttp://www.quantexpert.co.in/forum/viewtopic.php?f=10&t;=5&p;=574#p574
there also .... i have only posted got it from ...
https://www.facebook.com/wordpandit?ref=ts&fref;=ts
https://www.facebook.com/wordpandit?ref=ts&fref;=ts
@hatemonger said:U can choose any number from 1 to 96 .... if u choose 1 then n+1 will be 2 and n+2 will be 3 now if u choose 2 than u will have 3 numbers 2,3 and 4 product of this 3 number is divisible by 8 .... if u r choosing n as odd number than u wont have product divisible by except few cases such as 7 8 9 , 15 16 17 and so on ......even number to choose from 96 is 48 and 12 multiples of 8 so total 0f 60 numbers out of 96 .......
but will not 12 multiples of 8 will be in 48 even number