Official Quant thread for CAT 2013

MBA CAT 2022
3121
@sonamaries7 said:
A no N expressed to the base 5 is 232323...a total of 100 digits. What is the rem when N^4231 is divided by 4?0124
is it 1 sonamaries ;)
3122
@sonamaries7 said:
A no N expressed to the base 5 is 232323...a total of 100 digits. What is the rem when N^4231 is divided by 4?0124
@sonamaries7 said:
A no N expressed to the base 5 is 232323...a total of 100 digits. What is the rem when N^4231 is divided by 4?0124
2??
3123
@gs4890 said:
getting 10 sonamaries
logic??
3124
@sonamaries7 said:
A no N expressed to the base 5 is 232323...a total of 100 digits. What is the rem when N^4231 is divided by 4?0124
How come the option have 4
I'm getting 3.
Please someone clarify. :/
3125
@Brooklyn said:
2??
@Torque024 said:
How come the option have 4 I'm getting 3. Please someone clarify. :/
it is 0
3126
@sonamaries7 said:
it is 0
hw?? 😲
3127
@Brooklyn said:
hw??
thats for u guys to solve 😛 ...the soln isnt clear to me ...
3128
@sonamaries7 said:
thats for u guys to solve ...the soln isnt clear to me ...@krum : kahan hai!?
plz post d soln if u have 😃
3129
@Brooklyn said:
d
how ?
3130
@sonamaries7 said:
A no N expressed to the base 5 is 232323...a total of 100 digits. What is the rem when N^4231 is divided by 4?0124
when a number in base n is divided by n -1 , divisibility rule is sum of digits of N should be divisible by n -1

232323...a total of 100 digits --> sum is (2+3) * 50 = 250

250 in base 5 is 2000


2000 / 4 is perfectly divisible ..

so 0
3131
@sonamaries7 said:
A no N expressed to the base 5 is 232323...a total of 100 digits. What is the rem when N^4231 is divided by 4?0124
N = 232323....23 (100 digits) in base 5
= 2*5^99 + 3*5^98 + .... + 2*5 + 3

Now, 5 ż leaves remainder 1 when divided by 4

So, N will leave remainder 2 + 3 + 2 + 3 + .... + 2 + 3 (100 terms) = 5*50 = 250
means N will leave remainder 2 when divided by 4, hence N is an even number

(Even number)^4231 will be divisible by 4

So, remainder will be 0
3132
@gs4890 said:
how ?
Drop perpendicular to chord(OP) and calculate length OP. From OP you can get OM hence the ratio can be calculated
3133
@gautam22 said:
usko solve karoge end mein 48 aayega ...therefore 0
yar wat i did:

2*5^99 + 2*5^97 so on till 2*5^1 this give 2*50

now

3*5^98 + so on till 3*5^0 = 3*50
so 2*50+3*50 mod 4 = 2
3134
@chillfactor said:
N = 232323....23 (100 digits) in base 5= 2*5^99 + 3*5^98 + .... + 2*5 + 3Now, 5 ż leaves remainder 1 when divided by 4So, N will leave remainder 2 + 3 + 2 + 3 + .... + 2 + 3 (100 terms) = 5*50 = 250means N will leave remainder 2 when divided by 4, hence N is an even number(Even number)^4231 will be divisible by 4So, remainder will be 0
sir i got till 250

after dat, E(4)=2

so we get even number ^1 = 2
3135
@chillfactor said:
N = 232323....23 (100 digits) in base 5= 2*5^99 + 3*5^98 + .... + 2*5 + 3Now, 5 ż leaves remainder 1 when divided by 4So, N will leave remainder 2 + 3 + 2 + 3 + .... + 2 + 3 (100 terms) = 5*50 = 250means N will leave remainder 2 when divided by 4, hence N is an even number(Even number)^4231 will be divisible by 4So, remainder will be 0
(Even number)^4231 will be divisible by 4
(2)^4231 mod 4 =2 ?
Am I right sir?
3136
@gautam22 : no sir :nono:
3137
@naga25french said:
when a number in base n is divided by n -1 , divisibility rule is sum of digits of N should be divisible by n -1232323...a total of 100 digits --> sum is (2+3) * 50 = 250250 in base 5 is 20002000 / 4 is perfectly divisible ..so 0
I was applying same logic. But I added 2+5+0 =7 ; 7mod4 = 3; Where am I wrong sir?
Can you explain why have you done this? 250 in base 5 is 2000
3138
@naga25french said:
when a number in base n is divided by n -1 , divisibility rule is sum of digits of N should be divisible by n -1232323...a total of 100 digits --> sum is (2+3) * 50 = 250250 in base 5 is 20002000 / 4 is perfectly divisible ..so 0
Naga 250 is already a base 10 number, why are you converting it in base 5 to check the divisibility

Divisibility needs to checked always in base 10
@gautam22 said:
sir 2*5^99 + 3*5^98 + .... + 2*5 + 3....ko solve karenge to end mein 48 aayega.....aise bhi to ho sakta hai?
how did you get the last two digits as 48 ?? (I think its incorrect)

@Brooklyn said:sir i got till 250after dat, E(4)=2so we get even number ^1 = 2

You can't use Euler's if in a ż, a is not coprime to the divisor
3139
@Torque024 said:
(Even number)^4231 will be divisible by 4 (2)^4231 mod 4 =2 ? Am I right sir?
2^4231 (isn't it divisible by 4??? )
3140
@chillfactor said:
2^4231 (isn't it divisible by 4??? )
:P

Points P and Q lie on sides AC and BC of a ΔABC respectively with AB = 20 cm, BC = 12 cm and AC = 10 cm. PY is parallel to AQ and QX is parallel to BP such that Y lies on BC and X on AC. If A(ΔCXY) = 7.2 sq. cm, what is ℓ(XY)?
OPTIONS

1) 7.2 cm
2) 8 cm
3) 9.2 cm
4) 10 cm
5) Cannot be determined