The Official CAT 2007 Quant Thread

[krsh.vik]

Originally Posted by vineet.nitd

Y not sir

If the intermediate image to be formed lies on the 2nd axis that means the distance x between the source and the 1st axis is equal to the distance between the first axis and the intermediate image and hence the 2nd axes .Which means PM=PP=x+x=2x

yes it is ... misunderstood ..

[swatkat]

Originally Posted by mohit_353

3. How many numbers can be formed by using 8,7,5,3,2 such that they are divisible by

125 ? Digits to be used exactly once.

(1) 20 (2) 4 (3) 2 (4) 1

The answer is (2) ... 4 numbers can be formed by using 8,7,5,3,2.

[Destiny's_Child]

the positive integer n is such that if the first digit is moved to become the last digit then the new no is 7n/2..The no of digits in smallest n are...

I understood the approach that people have posted for this question but i was not able to figure out where i am going wrong

Heres my solution (following amar_kashyaps method of solving)

n=(10^m)a+b

7/2 (10^m a+b) = 10b + a
=> 7(10^m a + b) = 20b + 2a
=> 7.(10^m).a + 7.b =20b + 2a
=> 13b = a(7.10^m - 2)

now 10^m = 4 mod 13
10* 10^m = 40 mod 13 = 1
=> 10^(m+1) = 1 mod 13
=> m+1 = 12 (euler number for 13)

but i know its 6 and not 12......
whats my mistake????

[Destiny's_Child]

Originally Posted by vineet.nitd

Let the number N be xyz =11k => to be a multiple of 11 we must have x+z-y=11 m where , m = 0 or 1 (for a 3 digit number)
Case1:
m=0
hence k=10x+z = 2(x^2 +xz +z^2)
=>x^2+z^2+xz = 5x + z/2
Hence z shud be even .
For, z=0 , x=5 {discard x=0} giving N=550
For ,z=2 , x^2+3-3x=0 which does not have real roots and so will bge the case for z>2
Hence only one soln i.e N=550

case2 :
m=1
here min value of x and z must be 2 .
for x=2 , we can only have 209 (does nt satisfy)
for x=3 , we can have min val 308 , 3^2+8^2 >28 hence no other value also satisfies
for x=4 , 4^2+7^2 >37 hence no other gr8er value also satisfies
x=5 , , 5^2+6^2 >46 (No value satisfies)
x=6 , 5^2+6^2 > 55 (No value satisfies)
x=7 ,4^2+7^2 >64 (No value satisfies)
x=8 , 8^2+3^2=73 (satisfies) for 803 but no greater value than this
x=9 , 2^2 +9^2 >82 (No value satisfies)

Hence there are only 2 such numbers 550 and 803

I preetty much understood the case when m=0 but dint follow the case when m=1
Could someone please elaborate on the second case..........

[doomsayer]

Originally Posted by Destiny's_Child

the positive integer n is such that if the first digit is moved to become the last digit then the new no is 7n/2..The no of digits in smallest n are...

I understood the approach that people have posted for this question but i was not able to figure out where i am going wrong

Heres my solution (following amar_kashyaps method of solving)

n=(10^m)a+b

7/2 (10^m a+b) = 10b + a
=> 7(10^m a + b) = 20b + 2a
=> 7.(10^m).a + 7.b =20b + 2a
=> 13b = a(7.10^m - 2)

now 10^m = 4 mod 13
10* 10^m = 40 mod 13 = 1
=> 10^(m+1) = 1 mod 13
=> m+1 = 12 (euler number for 13)

but i know its 6 and not 12......
whats my mistake????

u r not wrong here....both 10^5 and 10^11 gives a reminder of 4 when divided by 13..
but the qustion is to find out the smallest number , so the answer is 5+1 =6....

[Destiny's_Child]

1. u have to weigh 1 to 40 kgs and u have to find the min number of weights required to do this

I find the weights required as 1 , 3, 9 ,27 ( minimum number of weights as 4) ......
Is that correct?????

@doomsayer
So that means i need to check all values of m for
10^m = 4 mod 13 where m<12 ?????

[doomsayer]

Originally Posted by Destiny's_Child

1. u have to weigh 1 to 40 kgs and u have to find the min number of weights required to do this

I find the weights required as 1 , 3, 9 ,27 ( minimum number of weights as 4) ......
Is that correct?????

@doomsayer
So that means i need to check all values of m for
10^m = 4 mod 13 where m<12 ?????

definitely,since the question is abt finding the smallest number....may be some value 10^100xxx might be satisfy the condition, but u cannt tell thats the ans...


regading weight wala queston ur ans is correct, In a normal thraaass where we can use both sides, by using those 4 weights we measure all weigts till 40

[Destiny's_Child]

I find it very difficult to solve problems of the type a^b^c^d^e

are there any particular steps tobe folllowed here to make them easy to crack????

For eg
What is the last digit of 22^33^44^55^66^77??


Also can somebody solve this problem

• A watch ticks 90 times in 95 seconds and another watch ticks 315 times in 323 seconds. If both the watches are started together, how many times will they tick in the first hour?
  A] 100 B] 101 C] 102 D] 99

I have a doubt as to the answer will be 100 and 101(i.e. its first together tick would be counted or not)

[doomsayer]

Originally Posted by Destiny's_Child

I find it very difficult to solve problems of the type a^b^c^d^e

are there any particular steps tobe folllowed here to make them easy to crack????

For eg
What is the last digit of 22^33^44^55^66^77??


Also can somebody solve this problem

• A watch ticks 90 times in 95 seconds and another watch ticks 315 times in 323 seconds. If both the watches are started together, how many times will they tick in the first hour?
  A] 100 B] 101 C] 102 D] 99

I have a doubt as to the answer will be 100 and 101(i.e. its first together tick would be counted or not)

last digit of 22^33^44^55^66^77 ---> 2 has a cylicity of 4 ,33^44^55^66^77 = 4k+1
>>>22^(4k+1)>>>>so the last digit is 2


ans of second questio is 101, u have to count their initial tick together

[Destiny's_Child]

Originally Posted by doomsayer

last digit of 22^33^44^55^66^77 ---> 2 has a cylicity of 4 ,33^44^55^66^77 = 4k+1
>>>22^(4k+1)>>>>so the last digit is 2

how do you come to arrive that
22^(4k+1)>>>>so the last digit is 2
agreed 33 = 4k + 1 but what about the other higher powers???