@Torque024 said:Follow your approach if n=1, that's each box will have only 1 ball
I thought n was the number of boxes. 
You're right.
You're right.
@Torque024 said:10 boxes are there each has n balls if n=9Make two set A(first 9 boxes),B(last box)From set A take 1 ball from first box 2 balls from 2nd box ... 9 balls form 9th boxIf first box contains 90grms balls then total weight(TW)=90+(2+3+4..9)*100And hence from the value of TW we can find then defective box in set A, and hence for totalif set A doesn't contains defective then B that's 10th ball will be defectiveSo weighting required will be 1If n=8Make two sets A(first 9 boxes),B(last 1 box)Apply same procedure for A to find defective box. Take out 0 balls from 1, 1 from 2nd and so onIf A don't have defective box, then B will have defective boxSo weighting required will be 1 editIf n=1Make 2 sets A(first 5) B(next5)Find total weight of A if !=5*100 then B will have defective box else AIn any case problem will reduce to 5 boxes1 weighting will reduce prblem to 5 boxes2 weighting will reduce to either 2 or 3Take worst 3now 2 weightings will be required total of 4 weightingsIf n=2 or If n=3EditMake 2 sets A(first 5) B(next5)Find total weight of A if !=5*100 then B will have defective box else AIn any case problem will reduce to 5 boxes1 weighting will reduce prblem to 5 boxes2 weighting will reduce to either 2 or 3Take worst 3from first take 1 ball, 2nd 2 balls and from 3rd three balls and measure the weight.For n=2 take out 0 balls 1st, 1 from 2nd 3 from 3rd and so on.Now here only 1 weighting will be required So total 3 weightings
when n=8
set A has 0 balls from 1st box , it is similar to set A(box 2 to box 9) and set B(first and last box)
if set A is not defective , how will you find if it is the first or the last box ??
so i think when n=8 , two weightings required .
set A has 0 balls from 1st box , it is similar to set A(box 2 to box 9) and set B(first and last box)
if set A is not defective , how will you find if it is the first or the last box ??
so i think when n=8 , two weightings required .
@rubikmath said:when n=8 set A has 0 balls from 1st box , it is similar to set A(box 2 to box 9) and set B(first and last box)if set A is not defective , how will you find if it is the first or the last box ??so i think when n=8 , two weightings required .
yes, edit kar dia :)
@gautam22 said:how many pairs of two digit numbers are there for whom last two digits of their squares are same?
13?
10,20..90
remember a square ends only in 1 4 5 6 9 0
square of an odd number always has even number of tens.
so 1 5 9 are not suitable in the case.
4 and 6 are units digits.
take 4
the numbers can be 12 22 32..92/18 28 38..98
observe that last 2 digits of a square of 2 digit number and square of its complement(diferenc on substraction from 100) are same.
ex: 12 &88
12^2=144
88^2=7744
so u dont need to calculate squares of bigger numbers while verifying whether its units and tens digit are same.
the numbers that satisfy are
12 38 and their complements( 88 62)
same is the case when the units digit is 6..
here u dont find any such numbers..
so total 13 such numbers
p.s:edited
10,20..90
remember a square ends only in 1 4 5 6 9 0
square of an odd number always has even number of tens.
so 1 5 9 are not suitable in the case.
4 and 6 are units digits.
take 4
the numbers can be 12 22 32..92/18 28 38..98
observe that last 2 digits of a square of 2 digit number and square of its complement(diferenc on substraction from 100) are same.
ex: 12 &88
12^2=144
88^2=7744
so u dont need to calculate squares of bigger numbers while verifying whether its units and tens digit are same.
the numbers that satisfy are
12 38 and their complements( 88 62)
same is the case when the units digit is 6..
here u dont find any such numbers..
so total 13 such numbers
p.s:edited
@gautam22 said:how many pairs of two digit numbers are there for whom last two digits of their squares are same?
Perfect squares end in 1,4,5,6,9,0
Ends in 00 -> All multiples of 10 => 10,20...90 (9 numbers)
Ends in 11 -> Either (x1)^2 or (x9)^2
(x1)^2 = 100x^2 + 20x + 1 (cannot end in 11, as tens digit is even)
(x9)^2 = 100x^2 + 180x + 81 (tens digit = 8 + 8x, cannot end in 11)
Ends in 44 -> Either (x2)^2 or (x8)^2
(x2)^2 = 100x^2 + 40x + 4 (tens digit = 4x , will end in 44, when x = 1,6)
(x8)^2 = 100x^2 + 160x + 64 (tens digit = 6 + 6x , will end in 44, when x = 3,8)
Ends in 55 -> (x5)^2
(x5)^2 = 100x^2 + 100x + 25 (always ends in 25)
Ends in 66 -> (x4)^2 or (x6)^2
(x4)^2 = 100x^2 + 80x + 16 (tens digit = 1 + 8x , cannot end in 66)
(x6)^2 = 100x^2 + 120x + 36 (tens digit = 3 + 2x, cannot end in 66)
Ends in 99 -> (x7)^2 or (x3)^2
(x3)^2 = 100x^2 + 60x + 9 (tens digit = 6x, cannot end in 99)
(x7)^2 = 100x^2 + 140x + 49 (tens digit = 4 + 4x, cannot end in 99)
So, final numbers are 10,20,30,40,50,60,70,80,90, 12, 62, 38, 88 [ 13 numbers ? ]
PS: Some results were standard, but no harm in seeing it for oneself :mg:
@gautam22
Question was.
How many pairs of two digit numbers are there for whom last two digits of their squares are same?
Last two digit is nothing but the N mod 100
10, 12, 20 I know only these that will satisfy our condition and I also know that
(50k+x)^2 and (50k-x)^2 will have last two digit as of x^2 coz' (50k-x)^2mod100 and (50k+x)^2mod100=x^2mod100
As x=10,12,20...I know only them
S={10,12,20}
Now we have to find the values of (50-x) or (50+x) repeatedly and add to the
satisfying set.
S={10,12,60,62,70,30,40,38}
Repeat
S={10, 12,20, 80, 90, 88, 60, 62, 70, 30, 40, 38}
Repeat
S={50, 10, 12, 20, 80, 90, 88, 60, 62, 70, 30, 40, 38}
Applying rule again will result in either number greater than 100 or the same number in the set.
So total of 13 numbers
Question was.
How many pairs of two digit numbers are there for whom last two digits of their squares are same?
Last two digit is nothing but the N mod 100
10, 12, 20 I know only these that will satisfy our condition and I also know that
(50k+x)^2 and (50k-x)^2 will have last two digit as of x^2 coz' (50k-x)^2mod100 and (50k+x)^2mod100=x^2mod100
As x=10,12,20...I know only them
S={10,12,20}
Now we have to find the values of (50-x) or (50+x) repeatedly and add to the
satisfying set.
S={10,12,60,62,70,30,40,38}
Repeat
S={10, 12,20, 80, 90, 88, 60, 62, 70, 30, 40, 38}
Repeat
S={50, 10, 12, 20, 80, 90, 88, 60, 62, 70, 30, 40, 38}
Applying rule again will result in either number greater than 100 or the same number in the set.
So total of 13 numbers
@gautam22 said:@Torque024@wovfactorAPS@soham2208 d way i did was like dis........difference between squares of 2 numbers can be written as ......(a-b)(a+b).....now this shud be equal to 100*n....for ex26-24)(26+24).....there can be many we just need to see...that a-b=k1....a+b=k2.....shud give value of a and b as integers(k1+k2=100n where n can vary from 1......99)........plz correct me if i m wrong
how is this related to ur question?
u are saying smthng like
a-b=12
a+b=42 on multiplication would give 504.
here last 2 digits of 504 are not same.
i think u misinterpreted the question
u are saying smthng like
a-b=12
a+b=42 on multiplication would give 504.
here last 2 digits of 504 are not same.
i think u misinterpreted the question
@IIM-A2013 said:@chillfactorit is 2.222%
I think they found the monthly interest rate as (80/3)/12 = 20/9 = 2.222 %
@wovfactorAPS said:how is this related to ur question?u are saying smthng likea-b=12a+b=42 on multiplication would give 504. here last 2 digits of 504 are not same.i think u misinterpreted the question
I think the question should be :-
How many two digit numbers are there such that last two digits of their square are same ??
OR How many two digits numbers 'ab' are there such that ab ˛ = _ _ xx (last two digits are same)
Q: in how many ways six diffenrnt balls can be put in 5 different boxes?
sabhi puys ko__/\__
@viewpt said:Q: in how many ways six diffenrnt balls can be put in 5 different boxes?sabhi puys ko__/\__
6^5 hai kya?
@viewpt said:Q: in how many ways six diffenrnt balls can be put in 5 different boxes?sabhi puys ko__/\__
15625
regards
scrabbler
@ankita14 said:6^5 hai kya?
5^6 hona chahiye na??
==================================================
Q: on a rectilinear path in same direction A is running at a speed of p=7.5 m/s and B is running at q=6.9 m/s where p>q and the distance b/w A and B =1mt. A is ahead of B. then wtithin how many seconds B will catch A?
==================================================
Q: on a rectilinear path in same direction A is running at a speed of p=7.5 m/s and B is running at q=6.9 m/s where p>q and the distance b/w A and B =1mt. A is ahead of B. then wtithin how many seconds B will catch A?
@viewpt said:Q: in how many ways six diffenrnt balls can be put in 5 different boxes?sabhi puys ko__/\__
Each ball 5 ways, Total 5^6 ways.
@viewpt said:5^6 hona chahiye na??==================================================Q: on a rectilinear path A is running at a speed of p=7.5 m/s and B is running at q=6.9 m/s where p>q and the distance b/w A and B =1mt. A is ahead of B. then wtithin how many seconds A will catch B?
If they are moving towards each other
Time A will catch B= Distance AB/(Sa+Sb)=1/14.4 sec
If they are moving in same direction
Time A will catch B, not possible.
Time A will catch B= Distance AB/(Sa+Sb)=1/14.4 sec
If they are moving in same direction
Time A will catch B, not possible.
@AIM_IIM_2013 said:6^5 hoga. isme jiske movement hoti h wo use k option dekhte h. 6 balls h har 1 k pas 5 options h.So 6^5 is the answer.@ankita14 aaj yaha ...
Bahi you said "6 balls h har 1 k pas 5 options h"
to 5*5*5*5*5*5 hoga na. 5^6
to 5*5*5*5*5*5 hoga na. 5^6
@viewpt said:5^6 hona chahiye na??==================================================Q: on a rectilinear path in same direction A is running at a speed of p=7.5 m/s and B is running at q=6.9 m/s where p>q and the distance b/w A and B =1mt. A is ahead of B. then wtithin how many seconds A will catch B?
Oh ha sorry :embarrassed:
@viewpt said:5^6 hona chahiye na??==================================================Q: on a rectilinear path in same direction A is running at a speed of p=7.5 m/s and B is running at q=6.9 m/s where p>q and the distance b/w A and B =1mt. A is ahead of B. then wtithin how many seconds A will catch B?
Is it A ahead or B?