@vijay_chandola said:log(a + b) c + logc (a + b)= [log {(a+b) }^2+ {log c ^2}]/{log(a+b) * log c}please provide options
try to do wdout options
How many ways to fill the 4X4 board by nonnegative integers, suchthat sum of the numbers of each row and each column is 3?a) 2006b) 2007
c) 2008
d) 2009
e) 2010
OA c)2008
c) 2008
d) 2009
e) 2010
OA c)2008
@krum @deedeedudu @staaalinnn @soumitrabengeri @scrabbler @grkkrg ...thoda prakash dalo yaar..
@staaalinnn said:Q> While adding all the page numbers of a book, I found the sum to be 1000. But then I realized that two page numbers (not necessarily consecutive) have not been counted. How many different pairs of two page numbers can be there?
24?
for 45 page numbers, we get the sum total to be 1035
But total = 1000
35 can be not be counted in the following 17 ways
1,34
2,33
3,32....and so on till 17,18
Next we can consider 46 pages
Total = 1081
81 can not be counted in the following 7 ways
34,47
35,46
36,45
37,44
38,43
39,42
40,41
After 46 if we consider 47..total comes out to be 1128 from which 128 cannot be subtracted as an addition of 2 numbers since the cap on the total number of pages is 46
So total pairs of numbers = 17+7 = 24
Please correct me if i am wrong
@staaalinnn said:Q> While adding all the page numbers of a book, I found the sum to be 1000. But then I realized that two page numbers (not necessarily consecutive) have not been counted. How many different pairs of two page numbers can be there?
n/2(2+(n-1)1)=1000
=>n(n+1)=2000
n=45=> s=1035
n=46=> s=1081
35=>17 pairs
81=>6 pairs
so 23 pairs?
edited
=>n(n+1)=2000
n=45=> s=1035
n=46=> s=1081
35=>17 pairs
81=>6 pairs
so 23 pairs?
edited
@krum said:n/2(2+(n-1)1)=1000=>n(n+1)=2000n=45=> s=1035n=46=> s=108135=>34 pairs81=>12 pairsso 46 pairs?
Bhai what you've done is not accounted for "different" pairs.. answer is 23 I guess..
@soumitrabengeri said:24?for 45 page numbers, we get the sum total to be 1035But total = 100035 can be not be counted in the following 17 ways1,342,333,32....and so on till 17,18Next we can consider 46 pagesTotal = 108181 can not be counted in the following 7 ways34,4735,4636,4537,4438,4339,4240,41After 46 if we consider 47..total comes out to be 1128 from which 128 cannot be subtracted as an addition of 2 numbers since the cap on the total number of pages is 46So total pairs of numbers = 17+7 = 24Please correct me if i am wrong
Its 23 pairs... (1,34),(2,33),...(17,18) ---- 17 pairs
(35,46),(36,45)...(40,41) ---- 6 pairs
An Easy one ..
@staaalinnn said:For divisibility by 9999: group into 4 digit numbers and add them and the sum should be divisible by 9999.Group the given number into 4 digits and addR[(4321 + 0432 + 1043 + 2104 + 3210) * 150]/9999=> R[(11110) * 150]/9999=> R[1111 * 150]/9999=> R[166650]/9999=> 6666 (Answer)
always nice to c good new concepts popping up .. that's makes this thread quite valuable :)
An extension of the above concept.. we check divisibility of 9 by grouping the given number into "any" group of numbers..
say for example.. 12345678910111213...................100 is to divisible by 9, we can group it as
say for example.. 12345678910111213...................100 is to divisible by 9, we can group it as
1|2|3|4|5|6|7|8|9|10|11|12|........|100| and add.. to get the sum ,then check the resulted number for divisibility..here (100*101)/2 = 50*101 =5*2=10 .. remainder is 1!
@soumitrabengeri said:24?for 45 page numbers, we get the sum total to be 1035But total = 100035 can be not be counted in the following 17 ways1,342,333,32....and so on till 17,18Next we can consider 46 pagesTotal = 108181 can not be counted in the following 7 ways34,4735,4636,4537,4438,4339,4240,41After 46 if we consider 47..total comes out to be 1128 from which 128 cannot be subtracted as an addition of 2 numbers since the cap on the total number of pages is 46So total pairs of numbers = 17+7 = 24Please correct me if i am wrong
34, 47 won't be considered bro.. since there were 46 pages at the 1st place
btw, superb approach! 
@albiesriram said:An extension of the above concept.. we check divisibility of 9 by grouping the given number into "any" group of numbers..say for example.. 12345678910111213...................100 is to divisible by 9, we can group it as 1|2|3|4|5|6|7|8|9|10|11|12|........|100| and add.. to get the sum ,then check the resulted number for divisibility..here (100*101)/2 = 50*101 =5*2=10 .. remainder is 1!
Extension of the above concept 😛 -
Quick divisibility rule:
Take 33 :
100/33 leaves remainder 1 . So one can frame divisibility rule of 33 as sum of digits taken two at a time from the right hand side of number .
Eg . Is 2112 divisible by 33 ?
12 + 21 = 33 . 33/33 perfectly divisible . Yes .. isn창€™t that easy ?
Third case:
Lets take an example of next number .. take 7
10/7 --> not divisible but remainder is not 0 or 1 or -1
100/7 --> not divisible but remainder is not 0 or 1 or -1
1000/7 --> not divisible but remainder is -1
so every three digit we will have -1 remainder ..
if 10^3 has -1 remainder , 10^6 has 1 as remainder...
so its alternative in nature for every three digits .
Divisibility rule of 7 will difference of sum of the digits taken at three at a time alternatively from RHS of the number and then check if this resultant divisible by 7
say number is 123130 to check for divisibility by 7 .,..
so according to this concept we have , 130 - 123= 7 which is divisible by 7
Or let창€™s say 7123130 -- > (130 + 7) 창€“ 123 = 14 which is divisible by 7
The beauty of this method is it works well for big number divisors (for instance- 143) .
235378/143 .. is it divisible ?
10^3 / 143 = remainder is -1
By this method ,we can say 378 - 235 = 143.. 143 / 143 is perfectly divisible .That's all you r done
Applying this formula , we can test divisibility of 2,3,4,5,7,8,9,11,27,13,33,37,99,77,143,259(37*7) etc ....
Courtesy - @naga25french
Quick divisibility rule:
Take 33 :
100/33 leaves remainder 1 . So one can frame divisibility rule of 33 as sum of digits taken two at a time from the right hand side of number .
Eg . Is 2112 divisible by 33 ?
12 + 21 = 33 . 33/33 perfectly divisible . Yes .. isn창€™t that easy ?
Third case:
Lets take an example of next number .. take 7
10/7 --> not divisible but remainder is not 0 or 1 or -1
100/7 --> not divisible but remainder is not 0 or 1 or -1
1000/7 --> not divisible but remainder is -1
so every three digit we will have -1 remainder ..
if 10^3 has -1 remainder , 10^6 has 1 as remainder...
so its alternative in nature for every three digits .
Divisibility rule of 7 will difference of sum of the digits taken at three at a time alternatively from RHS of the number and then check if this resultant divisible by 7
say number is 123130 to check for divisibility by 7 .,..
so according to this concept we have , 130 - 123= 7 which is divisible by 7
Or let창€™s say 7123130 -- > (130 + 7) 창€“ 123 = 14 which is divisible by 7
The beauty of this method is it works well for big number divisors (for instance- 143) .
235378/143 .. is it divisible ?
10^3 / 143 = remainder is -1
By this method ,we can say 378 - 235 = 143.. 143 / 143 is perfectly divisible .That's all you r done
Applying this formula , we can test divisibility of 2,3,4,5,7,8,9,11,27,13,33,37,99,77,143,259(37*7) etc ....
Courtesy - @naga25french
@Zedai said:34, 47 won't be considered bro.. since there were 46 pages at the 1st placebtw, superb approach!
oh yes..silly mistakes!!!! 
Thanks for pointing it out..
@gs4890 said:Extension of the above concept -Quick divisibility rule:Take 33 :100/33 leaves remainder 1 . So one can frame divisibility rule of 33 as sum of digits taken two at a time from the right hand side of number .Eg . Is 2112 divisible by 33 ?12 + 21 = 33 . 33/33 perfectly divisible . Yes .. isn't that easy ?Third case:Lets take an example of next number .. take 710/7 --> not divisible but remainder is not 0 or 1 or -1100/7 --> not divisible but remainder is not 0 or 1 or -11000/7 --> not divisible but remainder is -1so every three digit we will have -1 remainder ..if 10^3 has -1 remainder , 10^6 has 1 as remainder...so its alternative in nature for every three digits .Divisibility rule of 7 will difference of sum of the digits taken at three at a time alternatively from RHS of the number and then check if this resultant divisible by 7say number is 123130 to check for divisibility by 7 .,..so according to this concept we have , 130 - 123= 7 which is divisible by 7Or let's say 7123130 -- > (130 + 7) – 123 = 14 which is divisible by 7The beauty of this method is it works well for big number divisors (for instance- 143) .235378/143 .. is it divisible ?10^3 / 143 = remainder is -1By this method ,we can say 378 - 235 = 143.. 143 / 143 is perfectly divisible .That's all you r doneApplying this formula , we can test divisibility of 2,3,4,5,7,8,9,11,27,13,33,37,99,77,143,259(37*7) etc ....Courtesy - @naga25french
gr8 concept this.. Using this we we can find remainder of any number "n" in ANY BASE .all we have to find weather its leaving any +1 or -1 as remainder when (base)^k-1 is divided by n. k tends from 1 . Then we can frame our own divisibility rule..
EDITED
EDITED
Extension of the above concept -Quick divisibility rule:Take 33 :100/33 leaves remainder 1 . So one can frame divisibility rule of 33 as sum of digits taken two at a time from the right hand side of number .Eg . Is 2112 divisible by 33 ?12 + 21 = 33 . 33/33 perfectly divisible . Yes .. isn't that easy ?Third case:Lets take an example of next number .. take 710/7 --> not divisible but remainder is not 0 or 1 or -1100/7 --> not divisible but remainder is not 0 or 1 or -11000/7 --> not divisible but remainder is -1so every three digit we will have -1 remainder ..if 10^3 has -1 remainder , 10^6 has 1 as remainder...so its alternative in nature for every three digits .Divisibility rule of 7 will difference of sum of the digits taken at three at a time alternatively from RHS of the number and then check if this resultant divisible by 7say number is 123130 to check for divisibility by 7 .,..so according to this concept we have , 130 - 123= 7 which is divisible by 7Or let's say 7123130 -- > (130 + 7) – 123 = 14 which is divisible by 7The beauty of this method is it works well for big number divisors (for instance- 143) .235378/143 .. is it divisible ?10^3 / 143 = remainder is -1By this method ,we can say 378 - 235 = 143.. 143 / 143 is perfectly divisible .That's all you r doneApplying this formula , we can test divisibility of 2,3,4,5,7,8,9,11,27,13,33,37,99,77,143,259(37*7) etc ....Courtesy - @naga25french
Feels like am bestowed with some awesome sets of superpowers! 
Q> Let a, b, c, d be four real numbers such that a + b + c + d = 8 &
ab + ac + ad + bc + bd + cd = 12. Find the greatest possible value of d.
@staaalinnn said:Q> Let a, b, c, d be four real numbers such that a + b + c + d = 8 & ab + ac + ad + bc + bd + cd = 12. Find the greatest possible value of d.
bhai ek ganda sa ans lag rah hai..fir bhi is it 6?
P works faster than q. Will he take more than 20 days to complete a job X ?
1) The sum of times that p and q would take to complete X is 50 days
2) p would take 10 days less than q to complete x
opitons:
a)If the data in St1 alone is sufficient while data in st2 alone is not
b)data in st2 alone is suffice , while data in st1 alone is not
c)if the data in st1 alone or st2 alone but not both
d)if the data in both the st's together is suffice to answer the question
e)if the data even in both the st's together is not suffice to answer the question
1) The sum of times that p and q would take to complete X is 50 days
2) p would take 10 days less than q to complete x
opitons:
a)If the data in St1 alone is sufficient while data in st2 alone is not
b)data in st2 alone is suffice , while data in st1 alone is not
c)if the data in st1 alone or st2 alone but not both
d)if the data in both the st's together is suffice to answer the question
e)if the data even in both the st's together is not suffice to answer the question
Let p(x) = x^2 + bx + c , where b and c are integers. If p(x) is a
factor of both x^4 + 6x^2 + 25 and 3x^4 + 4x^2 + 28x + 5 , what is p(1)?
a) 0
factor of both x^4 + 6x^2 + 25 and 3x^4 + 4x^2 + 28x + 5 , what is p(1)?
a) 0
b) 4
c) 8
d) 12