Official Quant thread for CAT 2013

the_loser · January 20, 2013, 2:11pm

@getupsid said:
find remainder when 7+77+777+7777+......100 times is divided by 8?? find remainder when 2222^5555 and 5555^2222 is dividived by 7??

1 ) 6
2) zero.
if correct dan i wud try to explain, odrwise posting solution of no use.

ishu1991 · January 20, 2013, 2:12pm

@ishu1991
sry my bad its 6

the_loser · January 20, 2013, 2:12pm

@krum said:
P is a natural number.2P has 28 divisors and 3P has 30 divisors.how many divisors of 6P are there?P = 2^5*3^32P = 2^6*3^33P = 2^5*3^4so 6P = 2^6*3^4 hence 7*5=35 divisors

y p entails only 2 & 3?
i mean hw u identified P the exact?

getupsid · January 20, 2013, 2:14pm

@The_Loser said:
1 ) 62) zero.if correct dan i wud try to explain, odrwise posting solution of no use.

yes both are correct.......

enigma001 · January 20, 2013, 2:15pm

@getupsid said:
find remainder when 7+77+777+7777+......100 times is divided by 8?? find remainder when 2222^5555 and 5555^2222 is dividived by 7??

a) Remainder = 7+5+ 1*98 = 110 --> 6

b)Use Euler's Theorem .

The expression gets simplified as : 3^5 /7 = 5

The second expression gets simplified as : 4^2/7 = 2

estallar12 · January 20, 2013, 2:15pm

@getupsid said:
find remainder when 7+77+777+7777+......100 times is divided by 8?? find remainder when 2222^5555 and 5555^2222 is dividived by 7??

1.)

7 mod 8 = 7

77 mod 8 = 5

777 mod 8 = 1

7777 mod 8 = 1

...

So, The Series Reduces to => 7 + 5 +1 + 1 ... + 1 (98 times) mod 8

=> 12 + 98 mod 8 = 110 mod 8 = 6.

2.)

2222^5555 + 5555^2222 mod 7

2222^5555 mod 7 = 3^5555 mod 7.

3^3 mod 7 = -1

=> (3^3)^1851 mod 7 = -1

=> 3^5553 mod 7 = -1

So, 3^5555 mod 7 = -9 = 5.

5555^2222 mod 7 = 4^2222 mod 7.

We know, 4^3 mod 7 = 1

=> (4^3)^740 mod 7 = 1

=> 4^2220 mod 7 = 1

So, 4^2222 mod 7 = 16 = 2.

Thus, Overall 5+2 mod 7 = 0.

the_loser · January 20, 2013, 2:17pm

@The_Loser said:
1 ) 62) zero.if correct dan i wud try to explain, odrwise posting solution of no use.

Lo estallar sar ka explanation aa gya , the best one.

ravi.theja · January 20, 2013, 2:18pm

@getupsid said:
find remainder when 7+77+777+7777+......100 times is divided by 8?? find remainder when 2222^5555 and 5555^2222 is dividived by 7??

1. 7 + 5 + 1*98 / 8 = 110/8 = > rem =6.
2. zeroThe remainders when 5555 and 2222 are divided by 7 are 4 and 3 respectively. Hence, the problem reduces to finding the remainder when (4)^2222 + (3)^5555 is divided by 7.

Now (4)^2222 + (3)^5555 = (4^2)^1111 + (3^5)^1111 = (16)^1111 + (243)^1111. Now (16)^1111 + (243)^1111 is divisible by 16 + 243 or it is divisible by 259, which is a multiple of 7.

Hence the remainder when (5555)^2222 +(2222)^5555 is divided by 7 is zero.

mailtoankit · January 20, 2013, 2:21pm

@getupsid said:
find remainder when 7+77+777+7777+......100 times is divided by 8?? find remainder when 2222^5555 and 5555^2222 is dividived by 7??

6?

0?

krum · January 20, 2013, 2:21pm

@The_Loser said:
y p entails only 2 & 3?i mean hw u identified P the exact?

guess

getupsid · January 20, 2013, 2:24pm

1) find the last non zero digit of 96! ??

2) how many interesting digits are there??

ravi.theja · January 20, 2013, 2:25pm

@getupsid said:
1) find the last non zero digit of 96! ?? 2) how many interesting digits are there??

1.6 2.

96 = 5(19) + 1 = 2^19* 19! * 1! = 8*19! = 8*[ 5(3)+4 ] = 8* 2^3* 3!*4! = 8*8*6*4 = 6

5a+b = 2^a * a! *b! write in this format to get last non zero digit

bvdhananjay · January 20, 2013, 2:27pm

@Estallar12 said:
1.)7 mod 8 = 777 mod 8 = 5777 mod 8 = 17777 mod 8 = 1...So, The Series Reduces to => 7 + 5 +1 + 1 ... + 1 (98 times) mod 8=> 12 + 98 mod 8 = 110 mod 8 = 6.2.)2222^5555 + 5555^2222 mod 72222^5555 mod 7 = 3^5555 mod 7.3^3 mod 7 = -1=> (3^3)^1851 mod 7 = -1=> 3^5553 mod 7 = -1So, 3^5555 mod 7 = -9 = 5.5555^2222 mod 7 = 4^2222 mod 7.We know, 4^3 mod 7 = 1=> (4^3)^740 mod 7 = 1=> 4^2220 mod 7 = 1So, 4^2222 mod 7 = 16 = 2.Thus, Overall 5+2 mod 7 = 0.

estallar bhai eulers theorem kaise lagaya pls help.. Mjhe zyada kkbi smaj nai aaya kaise use krte h isse..

swapnil4ever2u · January 20, 2013, 2:31pm

@getupsid said:
1) find the last non zero digit of 96! ?? 2) how many interesting digits are there??

1)6

2)

cat.aspirant123 · January 20, 2013, 2:32pm

three unbiased dice are thrown together, probablity that all 3 show multiple of 3, but do not show the same multiple of 3 ?

estallar12 · January 20, 2013, 2:33pm

@getupsid said:
1) find the last non zero digit of 96! ?? 2) how many interesting digits are there??

1.)

Last non-zero digit of p! = Last non-zero digit of 4^n*(2n)! where n = P/10.

So, For 100!

=> Last Digit of = 4^10 * 20! = 6 * 4^2 * 4! = 6*16*24 = 96*24 = 4.

So, 100! = 100*99*98*97 * 96! = 4

=> Last Digit of 4 * 96! = 4

So, 96! Last Digit = 6. :)

2.) What's this ? 😲 :O

the_loser · January 20, 2013, 2:33pm

@ravi.theja said:
1.6 2. 96 = 5(19) + 1 = 2^19* 19! * 1! = 8*19! = 8*[ 5(3)+4 ] = 8* 2^3* 3!*4! = 8*8*6*4 = 65a+b = 2^a * a! *b! write in this format to get last non zero digit

kuch samaj nhi aa rha ya.

mailtoankit · January 20, 2013, 2:33pm

@getupsid said:
1) find the last non zero digit of 96! ?? 2) how many interesting digits are there??

1. 6?

2.

ravi.theja · January 20, 2013, 2:35pm

@Cat.Aspirant123 said:
three unbiased dice are thrown together, probablity that all 3 show multiple of 3, but do not show the same multiple of 3 ?

1/36?? "taking that all three cannot have same multiple ,but two can have same multiple"

estallar12 · January 20, 2013, 2:36pm

@bvdhananjay said:
estallar bhai eulers theorem kaise lagaya pls help.. Mjhe zyada kkbi smaj nai aaya kaise use krte h isse..

I have not used Euler's Theorem here.

It was just a Combination of Cyclicity of Powers of Numbers and Remainders.