@ vikky..wats gcd :confused: ..give me come clarity on this new conceptGCD - Greatest common divider 😃 same as HCF - Highest common factor
Cheers
MK
Nice thread..! :1eye:
Subscibing the thread....! :)
To start with... I have attempted a sum on some other thread..
it goes like.. find reminder when 1*1+11*11+111*111+.... (11... 1000times)*(1... 1000times) is divided by 100...
Some how I've found it must be 90.... just wanna verify, if i am going the right way.. as i attempted giving it a hit and trial approach.. !
Roll on puys..! 😃
Nice thread..! :1eye:
Subscibing the thread....! :)
To start with... I have attempted a sum on some other thread..
it goes like.. find reminder when 1*1+11*11+111*111+.... (11... 1000times)*(1... 1000times) is divided by 100...
Some how I've found it must be 90.... just wanna verify, if i am going the right way.. as i attempted giving it a hit and trial approach.. !
Roll on puys..! :)
The answer is 80.
For a detailed solution check out this post.
Kindly check out for the solutions on the thread where you found the post before asking for the solution again. There is a high probability that someone will have posted a detailed solution in there.
Here's a problem for starters.
A merchant mixes three varieties of rice (in integral amounts of weight in kgs) costing Rs.20/kg, Rs.24/kg and Rs.30/kg and sells the mixture at a profit of 20% at Rs.30 / kg. How many kgs of the second variety will be in the mixture if 2 kgs of the third variety is there in the mixture?
(1) 1 kg
(2) 5 kgs
(3) 3 kgs
(4) 6 kgs
The answer is 80.
For a detailed solution check out this post.
Kindly check out for the solutions on the thread where you found the post ........
Ok satan... agreed.. ! :thumbsup:
How about if we terminate the series at the 10th one...what should be the answer then..? I did it and found 90.. and generalized it for 1000 terms..! Is this not the right way..? Please do comment on this..!
@Satan:for rice mix problem....
ANS: 5
(Correct me if i am wrong)
let x,y,z be the kgs of rice resp.
so we have, 20x+24y+30(2) = 25(x+y+2)........(1)
(25 bcoz the rice is sold at 20% profit, real price is 25)
so,5x+y = 10....
y=10-5x.....only permissible value of x is 1, otherwise y will be come negative
(assuming x is int)
so y= 5.
Sri
Ok satan... agreed.. ! :thumbsup:
How about if we terminate the series at the 10th one...what should be the answer then..? I did it and found 90.. and generalized it for 1000 terms..! Is this not the right way..? Please do comment on this..!
If the same series is terminated on 10th term, the solution would become,
Remainder = (1 + 21*9) mod 100 = 190 mod 100 = 90
You can only generalize if you can prove that for the rest of the terms the remainder when divided by 100 is 0.
Here the rest of the terms give a remainder of = 21*990 mod 100 = -10*21 mod 100 = -10 mod 100 = 90
Now, since the remainder for rest of terms when divided by 100 is not 0, the remainder for whole series will become = Remainder(first 10 terms) + Remainder(remaining terms)= 90+90 = 180 mod 100 = 80
Hope that clarifies it.
Hi ..... I found this in numbers LOD2 ... can someone help me out with this one ....
How many zeroes will be there in
1^1 * 2^2 * 3 ^3 ...... 100^100 ?
Hi ..... I found this in numbers LOD2 ... can someone help me out with this one ....
How many zeroes will be there in
1^1 * 2^2 * 3 ^3 ...... 100^100 ?
I think the numbeer of 0 will be 1300, a 5 will provide on 0, do 5^5 will provide 5 zeros, for multiple of 25 there is one extra 5 so it will provide one extra zero.
so all 5^5 , 10^10.... will provide 5 + 10 + 20... 100 zeeros, there sum will come out to be, 1050 and then adding extra zeros due to on more multiple of 5 in 25, 50,75 ,100 = 250
hence total zeros will be 1300
is the answer 1220?
I think the numbeer of 0 will be 1300, a 5 will provide on 0, do 5^5 will provide 5 zeros, for multiple of 25 there is one extra 5 so it will provide one extra zero.
so all 5^5 , 10^10.... will provide 5 + 10 + 20... 100 zeeros, there sum will come out to be, 1050 and then adding extra zeros due to on more multiple of 5 in 25, 50,75 ,100 = 250
hence total zeros will be 1300
Din get what you are tryin to say .... can u please explain once more ASGHAN
I think the numbeer of 0 will be 1300, a 5 will provide on 0, do 5^5 will provide 5 zeros, for multiple of 25 there is one extra 5 so it will provide one extra zero.
so all 5^5 , 10^10.... will provide 5 + 10 + 20... 100 zeeros, there sum will come out to be, 1050 and then adding extra zeros due to on more multiple of 5 in 25, 50,75 ,100 = 250
hence total zeros will be 1300
Din get what you are tryin to say .... can u please explain once more ASGHAN
Sorry am bad at explaining 😃 , i think this question was solved in offical quant thread.
Anyways if the ans is right i can once again try to explain
yes dude the answer is write .....
I could only reach half the answer and your 5 Power funda i couldnt understand .... itll be great if u can re explain it or gimme the link to the post where the solution is given in the main thread
yes dude the answer is write .....
I could only reach half the answer and your 5 Power funda i couldnt understand .... itll be great if u can re explain it or gimme the link to the post where the solution is given in the main thread
i dont have the link will try to explain again
See a zero will come by the multiplication of 5*2, i.e one power of 5 and one power of 2 will give one zero, since the number of power of 5 is less that 2, so 5 will decide the max no. of 0
one power of 5 is giving one zero, so 2 power of 5 will give 2 0
i.e multiple of 5^2 =25 are 25 , 50 ,75, 100
so 5^5 will have 5 power of 5, and 25^25 will have 50 similarly 75^75 will have 75*2=150
so %^5 10^ 15^15 will form ap with first ter 5 and last term 100 and comman diff of 5, so add this ul get 1050 after adding, add 25 + 50+ 75 +100 once again as these are the multiple of 25, this comes out to be 250, hence 1050 + 250 =1300
hope its clear now
asghan .... 100 ^ 100 will give 200 zeroes and not 100 as u have stated .... i believe theres some flaw some where
10^10 20 ^20 ,.,..... etc etc
will give
10+20+30+40+50+60+70+80+90+200 = 650
with powers of 5 , there are 5+15..........+95 = 500 powers
so 650 +500 will give 1150 . Now if we are speaking about 25,75 etc , they are already covered in powers of 5 previously .... so i am a lil skeptical bout this ....
please clarify if i have a flawed understanding somewhere
got it !!! my own understanding mistake ! Thanks :rockon:
Hi
How do we proceed with problems in inequalities involving square root. Should we square and proceed.? For eg.
[ (x-2)/(1-2x) ]^1/2 > -1
Thanks
Hi
How do we proceed with problems in inequalities involving square root. Should we square and proceed.? For eg.
[ (x-2)/(1-2x) ]^1/2 > -1
Thanks
Yes u can square and proceed but b4 that u need to c d domain. for example here (x-2)/(1-2x) >=0 other wise lhs will be undefined. And this gives 1/2(x-2)>(1-2x) =>3x>3 =>x>1 . So our solution will be 1
Hursh wanted to subtract 5 from a number. Unfortunately, he added 5 instead of subtracting. Find the percentage change in result.
a) 300%
b) 66.66%
c) 50%
d) 33.33%
e)Can't be determined.
This might sound simple but please help me out...
Hursh wanted to subtract 5 from a number. Unfortunately, he added 5 instead of subtracting. Find the percentage change in result.
a) 300%
b) 66.66%
c) 50%
d) 33.33%
e)Can't be determined.
This might sound simple but please help me out...
in my opinion...we cannot determine d ans...
coz diff is 10 but original no is unknown...
pls correct me if am wrong...
in my opinion...we cannot determine d ans...
coz diff is 10 but original no is unknown...
pls correct me if am wrong...
Thnx bhavin...ya dats wat i thought....
but in d book its mentioned as 33.33%....so got confused...!!!
neway i think we cannot find a common answer bcoz the percentage will change with different no.s...