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Official Quant Thread for CAT 2011 [Part 8]

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A certain stock exchange designates each stock with one,two,three letter code,where each letter is selected from the 26 letters of the alphabet.if the letters may be repeated and if the same letters used in a different order constitute a different code,how manyt different stocks is it possible to uniquely designate with theses codes?
2951
8125
15600
16302
18278

The eternal seeker.....
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A thin wire 40 meters long is cut into two pieces.one piece is used to form a circle with radius r and the other is used to form a square.no wire is left over.which of the following represents a total area,in square meters,of the square and circular regions in tyerms of r?
Pi r^2
Pi r^2+10
Pi r^2+1/4pi^2 r^2
Pir^2+(40-2pi r)^2
Pi r^2+(10-1/2pi r)^2

The eternal seeker.....
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it should be 13.

no when divided by 5 will leave reminder as 3 and when by 3 it will leave reminder 1 so it is of the form 35K+13. so 13.

pls explain the bold part......

sorry typo. it is of the form 15K+13
see it is 5k+3 type as when it divided by 5 it will leave reminder 3, now 5k+3 should leave reminder 1 when divided by 3, so divide 5k+3 by 3 it will have reminder as 2K. now 2k should leave reminder 1 so smallest possible value of K is 2, and smallest no. is 5K+3 which is 13. so no. is of the form 15K (LCM of 5 & 3)+13
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Chintan Parikh, PR & Media Cell, IIT Kanpur Batch 2012-14
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it should be 13.

no when divided by 5 will leave reminder as 3 and when by 3 it will leave reminder 1 so it is of the form 35K+13. so 13.

pls explain the bold part......
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17 nov D Day Says
remainder when 7^15 divided by 15 ????

it should be 13.

no when divided by 5 will leave reminder as 3 and when by 3 it will leave reminder 1 so it is of the form 15K+13. so 13.

@ nachiketbhai CAT kaisi rahi
Chintan Parikh, PR & Media Cell, IIT Kanpur Batch 2012-14
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plz explain why the bold steps give different result or why in former m=8

See.. when we have 9*2*2; we have powers as (8, 1, 1)
But, power 8 can be taken from y or z. And other two will give (1, 1) as power. So, total 2 cases; one when y gives 8 and one when z gives 8.

When we have 4*3*3; powers are (3, 2, 2). Anyone can be 3 or 2.
So, we can have any one of x or y or z as power 3 and other 2 can take power 2. So, 3 cases. It is like 3 digit numbers with 2 digits being same.

When we have 6*3*2; powers are (5, 2, 1).
So, we have 3! ways.
It is like a digit number from 3 distinct digits.

Hope it is clear now..
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Is your question solved? You snipped it before I could clarify..

Which bold step? I think you missed making it bold.. Please clarify..

My take is 13

Euler's number for 15 is 8
So, 7^8k will leave remainder by 15 as 1
=> 7^16 will leave remainder as 1
Let, remainder by 7^15 be 'x'.

So, 7x will leave remainder by 15 as 1 (or -14)
=> x = -2 or 13

sorry just missed a bracket now corrected it
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tarun g Says
snipped...

Is your question solved? You snipped it before I could clarify..

fedbite Says
plz explain why the bold steps give different result or why in former m=8

Which bold step? I think you missed making it bold.. Please clarify..

17 nov D Day Says
remainder when 7^15 divided by 15 ????

My take is 13

Euler's number for 15 is 8
So, 7^8k will leave remainder by 15 as 1
=> 7^16 will leave remainder as 1
Let, remainder by 7^15 be 'x'.

So, 7x will leave remainder by 15 as 1 (or -14)
=> x = -2 or 13
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My take is 18

As a, b, and c are sqaures of primes number let those prime numbers be x, y and z resp.

So, N = x^6 * y^8 * z^10

We need 36 factors. And number can involve 2 or 3 prime numbers as with only one prime number we will need p^35 which is not possible.

Case 1: (p1)^m * (p2)^n
=> (m+1)*(n+1) = 36
So, (m, n) can be: (8, 3) or (5, 5) as m, n With (8, 3) we can have y or z as m and n can be other 2. So, total 2*2 = 4 cases
With (5, 5) we just need to pick 2 out (x, y, z) => 3C2 ways = 3 ways

So, total 7 ways for case 1

Case 2: (p1)^m * (p2)^n * (p3)^o ... it is 'o' and not '0' (zero)
=> (m+1)*(n+1)*(o+1) = 36
=> 9*2*2 or 6*3*2 or 4*3*3
9*2*2 => m = 8 => m can be from y or z => 2 ways
6*3*2 => m = 5 => m. n. o can be from any body => 3! = 6 ways
4*3*3 => m, n, o can be from anyone agian => 3!/2! = 3 ways
So, total 11 ways from case 2

In all; we have 7+11 = 18 ways

plz explain why the bold steps give different result or why in former m=8
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euler number of 15 is 8

so 7^15 mod 7 = 7^7 mod 7 = 49*49*49*7 mod 7 = 4*4*4*7 mod 7

= 1*28 mod 7 = 13

CAT-12: 98.94; CAT-11: 92.50 Calls: FMS, MDI, New IIM's (NITIE & IIFT not applied); Skipped: New IIM's; Converted: MDI PGPM 2015 batch
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17 nov D Day Says
sorry typo mistake its 7^15 is divided by 15..

(7^3)^5 mod 15 = (343)^5 mod 15 = (-2)^5 mod 15 = -32 mod 15 = -2 or 13..
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17 nov d day Says
remainder when 7^15 divided by 15 ????

=> (7^2)^7 * 7 % 15
=> 4^7 * 7 % 15
=> (4^2)^3 * 4 * 7 % 15
=> 1 * 4 * 7 % 15
=> 13
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tarun g Says
question is incomplete. 7^15 is divided by what?

sorry typo mistake its 7^15 is divided by 15..
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17 nov D Day Says
remainder when 7^15 divideb by????

question is incomplete. 7^15 is divided by what?
CAT-12: 98.94; CAT-11: 92.50 Calls: FMS, MDI, New IIM's (NITIE & IIFT not applied); Skipped: New IIM's; Converted: MDI PGPM 2015 batch
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snipped...

CAT-12: 98.94; CAT-11: 92.50 Calls: FMS, MDI, New IIM's (NITIE & IIFT not applied); Skipped: New IIM's; Converted: MDI PGPM 2015 batch
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