@Calvin4ever said:Is ur solution--> co-primes below 200 - co-primes below 100 ?
coprime below 100...
that is what is asked i guess...
13^73 + 14^3 MOD 11 =???
@Logrhythm said:coprime below 100...that is what is asked i guess...
What if the question was co-primes of 200 below 105, how wud u solve this ?? Forgive my dumbness 😛 :p
@Calvin4ever said:What if the question was co-primes of 200 below 105, how wud u solve this ?? Forgive my dumbness
then we wld count the co primes below 100 and count the primes between 100 and 105.. :)
@Logrhythm said:then we wld count the co primes below 100 and count the primes between 100 and 105..
@Calvin4ever said:What if the question was co-primes of 200 below 105, how wud u solve this ?? Forgive my dumbness
200 has 2 and 5 as prime divisors. So coprimes to 200 will also be coprime to 100.
Count no. coprimes to 100 and less than 100=100*(1-1/2)*(1-1/5)=40
Now count numbers between 101 and 105 which are coprime to 200.
They are 101,103.
So ans=40+2=42
ABCD a Convex quadrilateral . 3,4,5 and 6 points are marked on the sides AB,BC,CD,DA respectively . The No. of triangle with vertices on different side is..??
There are 100 players from 1 to 100 and 100 baskets from 1 to 100.
1st player puts 1 ball in every basket starting from the first baskt , second puts 2 in each an so on but starting from the second basket ( in baskets 2 , 4 , 6 ..)
Third player puts 3 in baskets in 3 , 6 , 9 .. ....
Process continues till 100th player puts 100 balls in 100th basket ..
Which basket wil have max no of balls
(a)96
(b)98
(c)100
(d)none f these
Please explain the approach too..
1st player puts 1 ball in every basket starting from the first baskt , second puts 2 in each an so on but starting from the second basket ( in baskets 2 , 4 , 6 ..)
Third player puts 3 in baskets in 3 , 6 , 9 .. ....
Process continues till 100th player puts 100 balls in 100th basket ..
Which basket wil have max no of balls
(a)96
(b)98
(c)100
(d)none f these
Please explain the approach too..
@bs0409 said:ABCD a Convex quadrilateral . 3,4,5 and 6 points are marked on the sides AB,BC,CD,DA respectively . The No. of triangle with vertices on different side is..??
pick one point from each side...
3*4*5 + 3*4*6 + 3*5*6 + 4*5*6 = 342...
Good night guys.. 
6 people share a pizza by cutting it out into 6 equal sectors . If 3 of them cut out and eat only the largest possible circle from their respective slices and leave the rest while others eat the whole slice then what is the percentage of pizza wasted?
(a)11
(b)15
(c)17
(d)22
(a)11
(b)15
(c)17
(d)22
@TootaHuaDil said:There are 100 players from 1 to 100 and 100 baskets from 1 to 100.1st player puts 1 ball in every basket starting from the first baskt , second puts 2 in each an so on but starting from the second basket ( in baskets 2 , 4 , 6 ..) Third player puts 3 in baskets in 3 , 6 , 9 .. .... Process continues till 100th player puts 100 balls in 100th basket ..Which basket wil have max no of balls(a)96 (b)98(c)100(d)none f thesePlease explain the approach too..
A basket will receive balls from palyers whose number is a factor of that basket number.
Like 6 has factors 1,2,3 and 6. So 6th basket will get balls from player no. 1,2,3,6.
So calculate no of factors of 96,98 and 100.
96: 2^5*3=12
98: 7^2*2=6
100: 2^2*5^2=9
So 96 will get max no. of balls
Like 6 has factors 1,2,3 and 6. So 6th basket will get balls from player no. 1,2,3,6.
So calculate no of factors of 96,98 and 100.
96: 2^5*3=12
98: 7^2*2=6
100: 2^2*5^2=9
So 96 will get max no. of balls
@TootaHuaDil said:There are 100 players from 1 to 100 and 100 baskets from 1 to 100.1st player puts 1 ball in every basket starting from the first baskt , second puts 2 in each an so on but starting from the second basket ( in baskets 2 , 4 , 6 ..) Third player puts 3 in baskets in 3 , 6 , 9 .. .... Process continues till 100th player puts 100 balls in 100th basket ..Which basket wil have max no of balls(a)96 (b)98(c)100(d)none f thesePlease explain the approach too..
this is problem based on factors...the number which has maximum number of factors will have the most number of balls..
96 has 12 factors...
now so does 60 = 2^2*3*5
so 96 hi hona chaie...confusion hai but...coz 12 factors max hai ye toh hai i guess but 60 ke bi toh hai 12 and 90 ke bhi...
i am half asleep :p
@Logrhythm said:this is problem based on factors...the number which has maximum number of factors will have the most number of balls..96 has 12 factors...now so does 60 = 2^2*3*5so 96 hi hona chaie...confusion hai but...coz 12 factors max hai ye toh hai i guess but 60 ke bi toh hai 12 and 90 ke bhi... i am half asleep
No, it shud be the one which has the maximum sum of factors and not the number of factors . Answer wud be 96 but still . . .
If S=[(104+324)*(224+324)*(344+324)*(464+324)*(584+324)]/[(44+324)*(164+324) *(284+324)*(404+324)*(524+324)].
Find S....any shortcut for this??
Find S....any shortcut for this??
@Calvin4ever said:No, it shud be the one which has the maximum sum of factors and not the number of factors .@bs0409
But Qn is "Which basket wil have max no of balls"
@Logrhythm said:1^2 = 12^2 = 43^2 = 94^2 = 65^2 = 56^6 = 67^2 = 98^2 = 49^2 = 110^2 = 0now we see only 1, 5, 6 and 0 end in the same digit as the number's unit digit...but 0 and 1 can be easily discarded...5 and 6 are left...xx6^2 always has a odd tens digit...we are left with 5...rest u can figure out easily which square it is... hope samajh aya hoga..
(376)^2 = 141376
How does your hypothesis justify this fact?
@bs0409 said:But Qn is "Which basket wil have max no of balls"
According to the question, the first guy puts 1 ball, the second guy puts 2 balls and so on , rite ? In case if only 1 ball was put each time, then the number of factors wud be enough!!