The Official CAT 2007 Quant Thread

[hanuman]

Originally Posted by mohit_353

another sitter....

1.)121212.......upto 300 digits is divided by 99, what is the

remainder ?
a)15
b)18
c)12
d)22

acc. to me ans. should be 18...not completely sure.....

wht to say???

1) 18 is the ans..
Split 99 as 9*11
121212.......upto 300 mod 9 is 0
In the options verify which of the option on mod 9 gives 0
here the options are too bad..if there was another option which is also a multiple of 9, then we may had to chinese remainder theorem

[horrible]

Originally Posted by mohit_353

so this means that if we change the question like if ten's digit is always greater than units digit or unit's digit is greater than ten's digit ...the answer will be same means 60.... .....

wht to say....

Indeed yes, if u take any one condition at a time...

[tatimatla]

another sitter....

1.)121212.......upto 300 digits is divided by 99, what is the

remainder ?
a)15
b)18
c)12
d)22

Can be solved by the Chinese Remainder method.

Let
121212...upto 300 digits = N

N mod 9 = 0 (sum of the digits)
N mod 11 = 150 mod 11 (sum of even digits - sum of odd digits) = 7

Hence N = 9A = 11B+7
A = (11B+7)/9

A should be an integer.
Easy to see that B = 1 clearly satisfies.

Hence, N = 11*9C + 11*1+7= 99C+18

NOTE: any number of the form 99C+18 is of the form 11B+7 and 9A.

[jaybind]

Originally Posted by tatimatla

another sitter....

1.)121212.......upto 300 digits is divided by 99, what is the


N mod 9 = 0 (sum of the digits)
N mod 11 = 150 mod 11 (sum of even digits - sum of odd digits) = 3

Hence N = 9A = 11B+7
A = (11B+7)/9

A should be an integer.
Easy to see that B = 1 clearly satisfies.

Hence, N = 11*9C + 11*1+7= 99C+18

NOTE: any number of the form 99C+18 is of the form 11B+7 and 9A.

can u make the highlighted parts more clear???i dint get it properly

[hanuman]

Originally Posted by mohit_353

2.)3) How many rectangles can one make from a regular chess board ?
a)1296
b)1332
c)1242
d)1221

Ans is b) 1332
No of rectangles including squares in chess board is 64*48/2=1536
No of squares in the chess board is 1^2+2^2+..8^2=204

hence no of rectangles in chess board is = 1536-204=1332

[doomsayer]

Originally Posted by jaybind

can u make the highlighted parts more clear???i dint get it properly

N mod 11 = 150 mod 11 (sum of even digits - sum of odd digits) = 3
>>>>>>>>>>>>Here i think what he meant is reminder is 7(not 3)

11*9C + 11*1+7= 99C+18
>>>>>>>
Here the first number which satisifies N = 9A = 11B+7 is 18,
that means this when div by 9 gives reminder 0 and when div by 11 give reminder 7.
so any number of the form 9*11k + 18 will satisfy the above condition..(div by 9 gives reminder 0 and when div by 11 give reminder 7.)

I hope its clear..

[doomsayer]

Originally Posted by tatimatla

another sitter....

1.)121212.......upto 300 digits is divided by 99, what is the

remainder ?
a)15
b)18
c)12
d)22

Can be solved by the Chinese Remainder method.

Let
121212...upto 300 digits = N

N mod 9 = 0 (sum of the digits)
N mod 11 = 150 mod 11 (sum of even digits - sum of odd digits) = 3

Hence N = 9A = 11B+7
A = (11B+7)/9

A should be an integer.
Easy to see that B = 1 clearly satisfies.

Hence, N = 11*9C + 11*1+7= 99C+18

NOTE: any number of the form 99C+18 is of the form 11B+7 and 9A.

another metod:
1212...= 12(100^149+100^148+.....................1)

which when div by 99 rem = 12*150>>>18

[..funky]

Originally Posted by tatimatla

another sitter....

1.)121212.......upto 300 digits is divided by 99, what is the

remainder ?
a)15
b)18
c)12
d)22

Can be solved by the Chinese Remainder method.

Let
121212...upto 300 digits = N

N mod 9 = 0 (sum of the digits)
N mod 11 = 150 mod 11 (sum of even digits - sum of odd digits) = 3

Hence N = 9A = 11B+7
A = (11B+7)/9

A should be an integer.
Easy to see that B = 1 clearly satisfies.

Hence, N = 11*9C + 11*1+7= 99C+18

NOTE: any number of the form 99C+18 is of the form 11B+7 and 9A.

rest im able to comprehend... except the part

• where the 7 came from... ?
• why did u take mod of 150 if the no. of digits are 300..
• how is the difference coming out to be 3

i know these may b elementary fundas... but my quants really sux... n need to work on it hard!!
And.if any 1 of u has time... could u present a more lucid explanataion...of Euler theorem... i think from the past year till now im pretty convinced that it plays a major major role in number system problems which are plenty in CAT!!

[tatimatla]

Originally Posted by ..funky

rest im able to comprehend... except the part

• where the 7 came from... ?
• why did u take mod of 150 if the no. of digits are 300..
• how is the difference coming out to be 3

Sorry for the typo...it's 7 and not 3...post edited.
The 150 came from the divisibility test for 11.
The number given was 1212...upto 300 digits
Now the divisibility test for 11 states that the remainder is the difference between sums of alternate digits. For 300 digits (even number) the remainder is the difference between sums of odd and even digits, and for an odd number of digits, the remainder is the difference between the sums of odd and even digits.
the given number has 150 2s and 150 1s...difference between even and odd digits = 150*2 - 150*1 = 150

Hope it's clear now...

And for those noobies who are looking to start off on the remainder probs, here's a headstart:
Khanna

Here's a more comprehensive link, with further updates on the discussions that took place last year:
Misc-Khullar

[rik_12]

Originally Posted by jaybind

any other methods for solving this???

divisors : 3 5

remainders: 1 3

subtracting: 2 2

LCM of 3 and 5 is 15.
Hence No is of the form 15K-2

hence remainder is (-2) i.e remainder :13