is the ans : 1000rs
@sahib_sheetu said: @YouMadFellow A SUM of money kept in bank amounts to Rs 1240 in 4 years and Rs 1600 in 10 years at simple interest. find the sum
@Enceladus said: 122333444455555...Till the single digit numbers we have 9*10/2 = 45 digits. Remaining digits = 1000 - 45 = 955.10 would occur 10 times.11 11 times, 12 12 times. and so on.=> 2*10 + 2*11 + 2*12 + 2*13 + .... + 2*99 = 2(10 + 11 + 12 + 13 + ... +n).At n = 32. this sum becomes 966.Hence, we have 966+45 = 1011 digits now. So last 4 digits would be 2323For remainder by 16, just consider the last 4 digits. => 2323 mod 16 = 3.
plz explain this que again..
from here -> 2*10+2*11+...+2*99...
from here -> 2*10+2*11+...+2*99...
@sahib_sheetu said: @Enceladustwo numbers a & b r such dat their GM is 20% lower than AM find ratio b/w numbrz. couldnt find the right ans
let the numbers be a and Ka... GM = a*sqrt(K); AM = a*(1+K)/2
GM = 0.8*AM ==> sqrt(K) = 0.4(1+K)
K = 4(1+K)^2/25
4+8K+4K2 = 25K
4K2 - 17K + 4 = 0
(4K-1)(K-4) = 0
K(ratio) = 1/4 or 4
Find the remainder when 21130 is divided by 1130
answer:
a. 1024
b.512
c.256
d.128
@[570309:thekrish] getting it 1120 .. Calculated by actual division ...
@thekrish said:Find the remainder when 21130 is divided by 1130 answer: a. 1024b.512c.256d.128
@junefever said:Ans a.1024
1130=113*10
2^1130 mod 113 gives remainder 7
2^1130 mod 10 gives remainder 4
113K+7=10p+4
113K+3=10p
for k=9 and p=102, we get the value as 1024.
hence, the answer is a.1024
@junefever said: 1130=113*10 2^1130 mod 113 gives remainder 7 2^1130 mod 10 gives remainder 4 113K+7=10p+4 113K+3=10p for k=9 and p=102, we get the value as 1024. hence, the answer is a.1024
can u xplain it elaborateldy please.... or can u give me the link where i can get
all the best ways to find out remainders ......
thanks
@thekrish said: Find the remainder when 21130 is divided by 1130 answer: a. 1024b.512c.256d.128
Plz check your Question it is coming 790......
@jain_ashu said: Answer should be 1. E(17) = 16, (38^16!)^1777 mod 17 = 38^16k mod 17 = 1
can you please tell me how E(17) = 16..? i dunno about this topic and i have just started with CAT preparations.... could someone help me out.. 😞
@thekrish said: Find the remainder when 21130 is divided by 1130 answer: a. 1024b.512c.256d.128
got answer as 790 by actual division....
@thekrish said: Find the remainder when 21130 is divided by 1130 answer: a. 1024b.512c.256d.128
I guess it should be 2^1130 mod 1130.
And these type of ques can be easily done with options.
1130 = 2*5*113.
2^1130 mod 2 = 0.
2^1130 mod 5 = 4.
Only option 1024 satifies. :D
hi..............please tell me which one is better to solve for NUMBERS
arun sharma or quantum cat (arihant publication). I have done time basic material for numbers but i think that those are very basic......and want to solve the problems that will appear in CAT..............please help
@LeoN88 said: Plz check your Question it is coming 790......
nope dude its 1024
only check again .once
and by the way can u tell me
a easier approach to find remainders for any problems such as this or any other kind
do we have any thread or link over here
@[597145:Rachanaramesh]...
let N = a^p * b^q
then E(N) = N*(1 - 1/a)*(1 - 1/b)
E(17)=17*(1 - 1/17)=16.
Hope it helps. It has been long time since I did this.
@[363542:rubber]... best place for numbers is totalgadha and PG. period. no need of books.
@[419941:Enceladus]: please explain ur method a bit more clearly
Can anyone correct me if i am wrong.
1.Arun Bikas and Chetakar have a total of 80 coins among them. Arun triples the number of coins with the others by giving them some coins from his own collection. Next Bikas repeats the same process. After this Bikas now has 20 coins. Find the number of coins he had at the beginning?
a. 11
b. 10
c. 9
d. 12
The ans i got was 20... but there is no such option only and the ans given in the book is 10
bro , answer comes down to be 20 for me too. that must be the answer