# Free Number Set I - 03 Practice Test - CAT

Consider four digit numbers for which the first two digits are equal and the last two digits are also equal. How many such numbers are perfect squares? (CAT 2007)

A. 2
B. 4
C. 0
D. 1
E. 3

#### SOLUTION

Solution : D

Soln:

Using the technique of finding squares of numbers upto 100 by keeping the base of 50 and 100( refer to the e-booklet for further details). We see that only squares of 12 will end with 144 (or the last 2 digits being the same). Hence we will have to find out if the first 2 digits will be the same for squares of 38, 62 (absolute difference of 12 from 50) and also for 88 (absolute difference of 12 from 100), we see that only 88 satisfies the condition, hence option 4.

The integers 1,2,...... 40 are written on a blackboard. The following operation is then repeated 39 times; in each repetition, any two numbers, say a and b, currently on the blackboard, are erased and a new number a+b-1 is written. What is the number left on the board at the end? (CAT 2007)

A. 820
B. 821
C. 781
D. 819
E. 780

#### SOLUTION

Solution : C

option (c) 781

Here, in each step we are adding two number and reducing the sum by 1. So after 39 operations, we will have the sum of all the numbers from 1 to 40 reduced by 39. Hence the final number will be S4039=781.

Three consecutive positive integers are raised to the first, second and third powers respectively and then added. The sum so obtained is a perfect square whose square root equals the total of the three original integers. Which of the following best describes the minimum value, say m, of these three integers?(CAT 2008)

A. 1<=m<=3
B. 4<=m<=6
C. 7
D. 10
E. 13<=m<=15

#### SOLUTION

Solution : A

Ans: a
It's an easy question. Trial with some numbers will give you the solution
1<=m<=3
31+42+53=144=122=(3+4+5)2.

What are the last two digits of 72008?   (CAT 2008)

A. 21
B. 61
C. 01
D. 41
E. 81

#### SOLUTION

Solution : C

Ans: (c) 01

The last two digits of a number is nothing but the remainder obtained when the number is divided by 100.This number leaves a remainder 1 when divided by 4 as well as 25. Hence the remainder obtained when this number is divided by 100 is also 1. Hence the last two digits of this number are 01.

Alternatively

Using Chinese remainder theorem: as we have to divide by 100 and find the reminder by 4 and then by 25 ,

i.e., 4A+1=25B+1, finding integer solutions we see that A=25 and B=4 will hence give a remainder of 01.

Alternatively, :

Use the last 2 digit rule for 7. Even here u will get 01 as the last 2 digits.

561 is divisible by : (CAT 1995)

A. 13
B. 31
C. 5
D. None of these

#### SOLUTION

Solution : B

Ans: (b)

(561)=(53)2(1)2=(125)2(1)2=(125+1)(1251)=126124=314126. It is therefore clear that the expression is divisible by 31.

For two positive integers a and b, define the function h (a,b) as the greatest common factor (GCF) of a, b. Let A be a set of n positive integers G(A), the GCF of the elements of set A is computed by repeatedly using the function h. The minimum number of times h is required to be used to compute G is : (CAT 1999)

A. n
B. (n - 1)
C. n
D. None of these

#### SOLUTION

Solution : B

Ans:

It is clear that for n positive integers function h (a,b) has to be used one time less than the number of integers, i.e., (n-1) times.

76n66n, where n is an integer >0, is divisible by (CAT 2002)

A. 13
B. 127
C. 559
D. All of these

#### SOLUTION

Solution : D

Ans: (d)

For n=1,7666=(73)2(63)2;

=(7363)(73+63)=(343216)(343+216);=127559=1271343. Hence it is divisible by 127,13,559.

When 2256 is divided by 17, the remainder would be : (CAT 2002)

A. 1
B. 16
C. 14
D. None of these

#### SOLUTION

Solution : A

Ans: (a)

2256 can be written as (24)64=(171)64.

In the expansion of (171)64every term is divisible by 17 except (1)64. Hence remainder is 1.

Or directly:

Euler's number of 17 is 16 and 256 is a multiple of 16, hence the remainder is 1.

What is the remainder when 496 is divided by 6? (CAT 2003)

A. 0
B. 2
C. 3
D. 4

#### SOLUTION

Solution : D

Ans: (d)

42=4 (mod 6)

44=4 (mod 6)

46=4 (mod 6)

and so on. The answer will remain the same.

The number of employees in Obelix Menhir Co. is a prime number and is less than 300. The ratio of the number of employees who are graduates and above, to that of employees who are not, can possibly be (CAT 2006)

A. 101 : 88
B. 87 : 100
C. 110 : 111
D. 85 : 98
E. 97 : 84

#### SOLUTION

Solution : E

Ans:(e)

Using options, we find that sum of numerator and denominator or 97 : 84 is (97 + 84) = 181 which is a prime number. Hence, it is the appropriate answer.

The remainder, when (1523+2323) is divided by 19, is : (CAT 2004)

A. 4
B. 15
C. 0
D. 18

#### SOLUTION

Solution : C

Ans: (c)

an+bn is always divisible by a+b when n is odd.

523+2323 is always divisible by 15+23=38. As 38 is a multiple of 19, 1523+2323 is divisible by 19.

Hence we get a remainder of 0.

992250 = 2a x 3b x 5c x 7d. The value of a+b+c+d is

A. 10
B. 9
C. 8
D. 7

#### SOLUTION

Solution : A

We know, 992250 = 21 x 34 x 53 x 72.

The value of a + b + c + d = 1 + 3 + 4 + 2 = 10

If x and y are integers then the equation 5x + 19y = 64 has (CAT 2003)

A. no solution for x<300 and y<0
B. no solution for x>250 and y>100
C. a solution for 250<x<300
D. a solution for 59<y<56

#### SOLUTION

Solution : C

Soln:

(c) 5x + 19y = 64

We see that if y = 1, we get an integer solution for x = 9, now if y changes (increases or decreases) by 5, x will change (decrease or increase) by 19.

Looking at options, if x = 256 we get y = 64.

Using these values we see option 1, 2 and 4 are eliminated and also that these exists a solution for 250<x<300.

For a positive integer n, let Pn denote the product of the digits of n and Sn denote the sum of the digits of n. The number of integers between 10 and 1000 for which Pn+Sn=n is (CAT 2005)

A. 81
B. 16
C. 18
D. 9

#### SOLUTION

Solution : A

Sol:

(a)10<n<1000

Let n is two digit number

n=10a+b ^aPn=ab,Sn=a+b

then, ab+a+b=10a+b

^aab=9a ^ab=9

There are 9 such numbers 19,29,33,....,99.

Then let n is three digit number

^a n=100a+10b+c ^a Pn=abc,Sn=a+b+c

Then abc+a+b+c=100a+10b+c

^aabc=99a+9b

^abc=99+9

But the maximum value for bc=81

And RHS is more than 99. Hence, no such number is possible.

Suppose the seed of any positive integer n is defined as follows:  Seed (n)=n, if n<10=seed(s(n)), otherwise,    Where s(n) indicated the sum of digits of n. For example, seed(7)=7, seed(248)=2+4+8=seed(14)=seed(1+4)=seed(5)=5, etc.   How many positive integers n, such that n

A. 39
B. 72
C. 81
D. 108
E. 55

#### SOLUTION

Solution : E

Sol:

e. 55

From the definition of "seed"?, it is clear that we have to count number of integers between 1 and 500, which are divisible by 9. The smallest is 9 and the largest is 495. 9 A__1=9 and 9 A__55=495. Hence there are 55 such numbers.