Official Quant thread for CAT 2013

MBA CAT 2024
24268
@iLoveTorres said:
@amresh_mavericki think it is 108
Again, I am not sure....since 999 is divisible by 27 while 121212....is not, the answer should not be a multiple of 27....works out to 27k+18 form I think....isliye answers pooch raha tha, too lazy to do the remaining...could be wrong though :P

regards
scrabbler
24269
@scrabbler said:
Don't think it is 112 as 121212....300 digits is divisible by 9 and so is 999, so the remainder also should be a multiple of 9Sirji options kya they original mein?regardsscrabbler
my bad.... i did for 121121.... i guess answer should be 123. what say?
24270
@amresh_maverick 1) 11 2) 666 3) no is of form x36 so in total 9 such no.s
24271
@techgeek2050 said:
my bad.... i did for 121121.... i guess answer should be 123. what say?
Not multiple of 9 :(

regards
scrabbler
24272
@scrabbler said:
Not multiple of 9 regardsscrabbler
@amresh_maverick 594 it is i suppose, though i think that 1212121212.... is divisible by 27
24273
@amresh_maverick said:
2>1212121212......................... 300 digits when divided by 999 rem ?
Duh....is it 666?

If yes, then I am the epitome of stupid :(

regards
scrabbler
24274
@scrabbler said:
Not multiple of 9 regardsscrabbler
mera toh answer hi 18 aa rha h
24275
@scrabbler said:
Duh....is it 666?If yes, then I am the epitome of stupid regardsscrabbler
oh yes... 666 aa rha hai
24276
@scrabbler said:
Not multiple of 9 regardsscrabbler
324 ?
24277

999 se chote saare multiples of 9 post hone waale h aaj yahan

24278
@Narci said:
oh yes... 666 aa rha hai
Approach bata bhai??

24279
@scrabbler said:
Duh....is it 666?If yes, then I am the epitome of stupid regardsscrabbler
@amresh_maverick nahi bahi its 333 i m dead sure now :P
24280
@scrabbler said:
Duh....is it 666?If yes, then I am the epitome of stupid regardsscrabbler
Approach bata bhai??
24281
@scrabbler said:
Duh....is it 666?If yes, then I am the epitome of stupid regardsscrabbler
aap aur galat , ho nahi sakta (2 times)

A no N^2 is such that exactly 3 of its factors are less than N. The no of factors of N^3 ?


24282
@amresh_maverick said:
1> 2^469 + 3^268 when divided by 22 rem ?2>1212121212......................... 300 digits when divided by 999 rem ?3>How many 3 digit nos exist , the cubes of which end in 56 ?
OA:
1> 11
2> 666
3> 18

Details soln refer posts by @scrabbler sir
24283
@amresh_maverick said:
aap aur galat , ho nahi sakta (2 times)A no N^2 is such that exactly 3 of its factors are less than N. The no of factors of N^3 ?
10 ?
24284
@Narci said:
10 ?
correct , approach plz
24285
@sachisurbhi said:
find the remainder when 84^79 is divided by 100a)24b)44c)64d)84
OA) 64
24286
@amresh_maverick said:
correct , approach plz
for any number N which has some number of factors..
its square N^2 will have N as the middle factor if we arrange all the factors of N^2 in an order..

so if N^2 has 3 factors less than N, therefore N^2 has 7 factors...

thus N^2 = a^6..

therefore N^3 = a^9...

thus 10 factors..
24287
@saurav205 said:
Approach bata bhai??
OK so bear with me, this is gonna be a bit long…

Consider a number 123456. Let's try and get the remainder with 999.

Let's split it as 123000 + 456.

Now 123000 = 123 *1000 = 123 * (999+1) = 123*999 + 123. So remainder with 999 is 123.

Also 456 ka remainder with 999 is of course 456. So total remainder is equivalent to 123 + 456 = 579.

Similarly if we try to find the remainder of 123234345 with 999 we can show it will be ≡ 123 + 234 + 345 = 702 and so on.

So if we want to find the remainder of a number with 999, split it into groups of 3 digits (starting from the last) and add them.

So in this case 1212121212121212….300 digits ≡ 121212 50 times ≡ (121 + 212) + (121 + 212)…50 times which is ≡ 333 * 50 when divided by 999.

Now Rem(333 * 50) /999 ≡ 333 *Rem (50/3) = 333 * 2 = 666.

And there I was trying to apply Chinese Remainder and whatnot 😞 😞

@iLoveTorres
@Narci
@techgeek2050
@amresh_maverick

regards
scrabbler