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5.What is the remainder when 123123123.....(300 digits) is divided by 99

(a)18 (b)27 (c)33 (d)36 (e)39

Can be written as 123*100^150 + 123*10*100^148 + 123*100^147 + 123*10*100^145..........123*100^3 + 123*10*100^1+123..

i.e taking rem(100/99)=1..we get it 123*(1+1+1....50times)+123*10*(1+1..50times)

ie...123*50+123*10*50+123=rem(123*551)

Am not getting the answer !!!!

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@seetharam7 said:

5.What is the remainder when 123123123.....(300 digits) is divided by 99

(a)18 (b)27 (c)33 (d)36 (e)39

Find the remainder seperately when it is divided by 9 and 11.
By 11 ull get 0. As the sum of odd digits = sum of even digits.
With 9 ull get 6. since the sum of digits is 600.
11K = 9R + 6.
when R = 3, K =3.
So remainder is 11*3 = 33
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@raku1989 Thanks.
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Can someone post the solution to the 1st and 2nd ques ?

845
@Kevin88 said: 1)If a, b, c, d are real numbers with a2 + b2 + c2 + d2 = 100, then what is the maximum value of 2a + 3b + 6c + 24d?
a)240 b)250 c)300 d)None of these
This one can be done this way....
2a+ 3b + 6c + 24d will be max when
a/2 = b/3 = c/6 = d/24
Substitute this in eqtn a^2 +b^2+c^2+d^2=100
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Ull get a= 20/25 b = 30/25 c= 60/25 and d= 240/25

So 2a+ 3b + 6c + 24d = (40+90+360+5760)/25

= 6250/25 = 250.

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@seetharam7 said:Can someone post the solution to the 1st and 2nd ques ?
Qn 1) a^n- b^n will be divisible byboth (a-b) and (a+b) when n is even.
so 18^2000-1 will be divisible by 17 and 19. and 12^2000-5^2000 will be divisible by 17 and 7
Here 17 is the common divisor
Do the same for next combination(18 and 5. 12 and 1)ull get 13 as thge common divisor.
So that means the whole term is divisible by 17*13 = 221
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@seetharam7 said:Can someone post the solution to the 1st and 2nd ques ?
If n = 200, S=8^200.5^600 = 10^600. It will have 601 digits starting with 1 and 600 zeros.
If n = 201 ,it will start will 8 and 600 zeros. So here u just have to find out the least power of 8 which has four digits. Ie 8^4 = 4096.
So if n=204 it will start with 4096 and 600 digits. (so a total of 604 digits)
Sum of digits will be 4+0+9+6 = 19
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@raku1989 Your answers are perfect as always buddy 😃 As for the binomeal theorem problem, an alternative approach is

(1+2x+3x^2-4x^3)^10 = ao+a1x+a2x^2+...........

Putting x=1 on both sides, a0+a1+a2+a3+............. = (1+2+3-4)^10 = 1024

850
1)Allwin went to the market and bought some chikoos, mangoes, and bananas. Allwin bought 42 fruits in all. The number of bananas is less than half the number of chikoos; the number of mangoes is more than one-third the number of chikoos and the number of mangoes is less than three-fourths the number of bananas.
How many more/less bananas did Allwin buy than mangoes?
1)3 2)14 3)6 4)11 5)None of these
851

2) The set S has 5 elements. In how many ways can one select two (possibly identical) subsets of S whose union is S?

1)32 2)31 3)64 4)128 5)122
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@raku1989 said:
1)Allwin went to the market and bought some chikoos, mangoes, and bananas. Allwin bought 42 fruits in all. The number of bananas is less than half the number of chikoos; the number of mangoes is more than one-third the number of chikoos and the number of mangoes is less than three-fourths the number of bananas.
How many more/less bananas did Allwin buy than mangoes?
1)3 2)14 3)6 4)11 5)None of these
The ans is 3 ryt ? I used trial and error to get the number of chikoos,bananas and mangoes as 23,11 and 8 respectively
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@
@raku1989 For question 2 the answer is 122 ?
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@rohan_bhasker said:
The ans is 3 ryt ? I used trial and error to get the number of chikoos,bananas and mangoes as 23,11 and 8 respectively
Yup....I too did the same...
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@rohan_bhasker said: @raku1989
@raku1989 For question 2 the answer is 122 ?
Yup ans is 122....
I din get this....later i read the soln provided....
Which way you did?
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@raku1989 i did it in a random manner and found out that 64+ possibilities exist.Then between the other two options, i decided to take a guess.Even i found the solution provided on the prep thread to be far more logical
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Does anyone here take TIME's MCs and CRTs ?

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@seetharam7 I haven't started yet.will start this month.........
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Hey, whats everyone doing ? This thread has been quiet. Whats your study plan for the last month ?

Found this revision thread
http://www.pagalguy.com/forums/cat-and-related-bschools/cat-revision-topic-day-t-85202/p-3601173?page=1


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There are 90 questions in a test. Each correct answer fetches 1 mark, each wrong answer attracts a penalty of 1/4 mark and each unanswered question attracts a penalty 1/8 mark. If a candidate scored 23 marks, what is the minimum no of questions answered wrongly by him ?


4, 7 ,12, 5

???