Official Quant thread for CAT 2013

MBA CAT 2024
24888
@iLoveTorres said:
do you mind explaining this part
yes sure,
a ka range h wo. 33.
like a=0, will give u the first pair which isnt divisible by 9. 1^2*2^3
then a= 1 will give.. 4^2*5^3
...
finally.. a=32 ... 97^2*98^3.

so total 33 such terms. i.e a=0 to a=32.
got it?
24889
What are the last three digits of the greatest common divisor of 2^(2^100−1) − 1 and 2^(2^105−1) − 1 ?
24890
@jain4444 said:
What are the last three digits of the greatest common divisor of 2^(2^100−1) − 1 and 2^(2^105−1) − 1 ?
is it 1??

bohot hi bekaar method use kara hai waise meine......but if my method is correct then it proves that these 2 numbers are co-prime to each other....but wo prove kaise karu wo samajh nahi aa raha...
24891
@ChirpiBird said:
yes sure, a ka range h wo. 33. like a=0, will give u the first pair which isnt divisible by 9. 1^2*2^3 then a= 1 will give.. 4^2*5^3...finally.. a=32 ... 97^2*98^3.so total 33 such terms. i.e a=0 to a=32. got it?
what about the last term? 99^2*100^3?
24892
@iLoveTorres said:
what about the last term? 99^2*100^3?
Arrey..those terms are the terms which are not divisible by 9.
Last term nahi ayega na..it is not of the form (3k+1)^2*(3k+2)^3
24893
@jain4444 said:
What are the last three digits of the greatest common divisor of 2^(2^100−1) − 1 and 2^(2^105−1) − 1 ?
647?
24894

Most of you seem to have reached some advanced level :O
Guys how many have just started (less than 1 month )preparations for CAT-2013 ?

24895
@ScareCrow28 said:
Arrey..those terms are the terms which are not divisible by 9.Last term nahi ayega na..it is not of the form (3k+1)^2*(3k+2)^3
got it bro.. thanks :)
24896
@jain4444 said:
What are the last three digits of the greatest common divisor of 2^(2^100−1) − 1 and 2^(2^105−1) − 1 ?
if i am not wrong the approach should be 2^x-1 , 2^y-1
so the gcd wil be 2^(gcd[x,y])-1
now find the gcd of 2^100-1 and 2^105-1 here again applying the same rule you will get gcd as 2^5-1=31
hence the gcd of 2^x-1 and 2^y-1 will be 2^31-1
now to find the last three digits divide by 1000 and find it :P
24897
@Logrhythm said:
is it 1?? bohot hi bekaar method use kara hai waise meine......but if my method is correct then it proves that these 2 numbers are co-prime to each other....but wo prove kaise karu wo samajh nahi aa raha...
Nahin not co-prime...kuch to ganda sa hai...295 types hai shayad...

regards
scrabbler
24898
@iLoveTorres said:
what about the last term? 99^2*100^3?
this is divisible by 9 na... 9*11 = 99 :)
24899
@jain4444 said:
What are the last three digits of the greatest common divisor of 2^(2^100−1) − 1 and 2^(2^105−1) − 1 ?
let me share my ghatiya approach...i found out the last 3 digits individually and then the GCD..

2^(2^100 - 1) - 1 mod 1000

1000 = 125*8

2^(2^100 - 1) - 1 mod 8 = 7

2^(2^100 - 1) - 1 mod 125

2^(2^100 - 1) mod 125

2^100 - 1 mod 100 = 75

2^75 - 1 mod 125 = 67

8p+7 = 125q+67 --> 567

2^(2^105 - 1) - 1 mod 1000

2^(2^105 - 1) - 1 mod 8 = 7

2^(2^105 - 1) - 1 mod 125

2^105 - 1 mod 100 = 31

2^31 - 1 mod 125 = 22

8x+7 = 125y+22 --> 647

GCD(567,647) = 001

abb pls bata do kya galti kar raha hun mein...jisse mein apne muu par anda maar saku

@scrabbler dekho ek baar...
24900
@Logrhythm said:
let me share my ghatiya approach...i found out the last 3 digits individually and then the GCD.. 8p+7 = 125q+67 --> 5678x+7 = 125y+22 --> 647GCD(567,647) = 001abb pls bata do kya galti kar raha hun mein...jisse mein apne muu par anda maar saku
GCD of last 3 digits != GCD of the numbers themselves...

For example, 1331 and 121 ka GCD is 121 but GCD (121, 331) = 1.

regards
scrabbler
24901
@Logrhythm said:
let me share my ghatiya approach...i found out the last 3 digits individually and then the GCD.. 2^(2^100 - 1) - 1 mod 1000 1000 = 125*82^(2^100 - 1) - 1 mod 8 = 72^(2^100 - 1) - 1 mod 125 2^(2^100 - 1) mod 125 2^100 - 1 mod 100 = 752^75 - 1 mod 125 = 67 8p+7 = 125q+67 --> 5672^(2^105 - 1) - 1 mod 10002^(2^105 - 1) - 1 mod 8 = 72^(2^105 - 1) - 1 mod 1252^105 - 1 mod 100 = 312^31 - 1 mod 125 = 228x+7 = 125y+22 --> 647GCD(567,647) = 001abb pls bata do kya galti kar raha hun mein...jisse mein apne muu par anda maar saku @scrabbler dekho ek baar...
Yaar tumne last 3 digits ka gcd kyu nikala? It is not same as gcd of whole number na? Aise thodi kar sake ho 😛 ya kar sakte ho?
24902
@ScareCrow28 said:
Yaar tumne last 3 digits ka gcd kyu nikala? It is not same as gcd of whole number na? Aise thodi kar sake ho ya kar sakte ho?
sirf mein kar sakta hun....coz mujhe -1 lene ka bohot shauk hai...aap mat karna...
24903
@iLoveTorres said:
if i am not wrong the approach should be 2^x-1 , 2^y-1so the gcd wil be 2^(gcd[x,y])-1now find the gcd of 2^100-1 and 2^105-1 here again applying the same rule you will get gcd as 2^5-1=31hence the gcd of 2^x-1 and 2^y-1 will be 2^31-1now to find the last three digits divide by 1000 and find it



answer is 647
24904
@Logrhythm said:
sirf mein kar sakta hun....coz mujhe -1 lene ka bohot shauk hai...aap mat karna...
kuch shauk ache hote hai :P
24905
@iLoveTorres said:
if i am not wrong the approach should be 2^x-1 , 2^y-1so the gcd wil be 2^(gcd[x,y])-1now find the gcd of 2^100-1 and 2^105-1 here again applying the same rule you will get gcd as 2^5-1=31hence the gcd of 2^x-1 and 2^y-1 will be 2^31-1now to find the last three digits divide by 1000 and find it
Seriously had no idea of this before :splat: that was cool..
24906
@ScareCrow28 said:
Seriously had no idea of this before that was cool..
and i am sure i wld say the same thing every time someone posts this shortcut.....coz mujhe yaad nahi hote ye sabb...yaar :(
24907
41!^2 find remainder by 83 ?