# Free Numbers Set II - 01 Practice Test - CAT

### Question 1

N= 4! + 5!+6!.........200!. Find the last two digits of N

#### SOLUTION

Solution :C

After 10!, the factorials end with 0 as the last two digits. We thusneed to find the last two digits from 4! To 9! Only

4!+5!+6!....9!= 24+120 + 720 + 5040 + 40320 + 362880 = ___04

### Question 2

Find the sum of the series N=13+115+135.........100 terms

#### SOLUTION

Solution :C

First term of the series (13)=1(3×1)=12[1−13]

Second term of the series (115)=1(3×5)=12[13−15]

Third term of the series (135)=1(7×5)=12[15−17]

..............................................................................

100th term of the series (1199×201)=12[1199−1201]

12[1−13+13−15+15−17+17...................+1199−1201]

12[1−1201]=200(2×201)=100201

### Question 3

Find the 202nd digit from the right in the product of 4!×5!×6!...............71!?

#### SOLUTION

Solution :D

Note that 71! has 16 zeroes at its end.(You can find this out by finding the highest power of 5 in 71!)

So does 70!.

66! To 69! Have 15 zeroes at its end. Multiplication of these itself amounts to >202

Hence the 202nd digit is 0

To find the highest power of a number in a factorial

a)Highest power of a prime number in a factorial:

To find the highest power of a prime number (x) in a factorial (N!), continuously divide N by x and add all the quotients.

Eg) Find the highest power of 100!

Solution:

1005=20;205=4;

Adding the quotients, its 20+4=24. So highest power of 5 in 100! = 24

b)Highest number of a composite number in factorial

1)Factorize the number into primes.

2)Find the highest power of all the prime numbers in that factorial using the previous method.

3)Take the least power.

### Question 4

If (127)8+(12)6=(x)5. Find the value of x.

#### SOLUTION

Solution :C

Write all the numbers in base 10 and then arrive at the answer

127 in base 8=7×1+2×8+1×82=87 in base 10

12 in base 6=2×1+1×6=8 in base 10

Adding both =95 in base 10

Expressing in base 5=340

### Question 5

What is the remainder when (32540×42400×3233)+222 is divided by 13?

#### SOLUTION

Solution :C

32540 using the divisibility test for 13 ( triplets at odd- triplets at even places ) can be written as

540-32= 508. 508 when divided by 13 can be represented as 13k + 1 (i.e. when the number is divided by 13, the remainder is 1)

Similarly

42400= 400-42= 358 = 13k + 7

3233= 233-3=230= 13k + 9

(13k + 1)( 13k + 7)(13k + 9) will give the non-divisible part as (9×7×1)=63

therefore 63+222= remainder of 285 when divided by 13= -1 or 12

### Question 6

How many natural numbers between 1 and 900 are NOT multiples of any of the numbers 2, 3, or 5?

#### SOLUTION

Solution :A

Note: This is based on the concept of Euler’s number

A number N can be written as ambn. In this case, N=900=22×32×52

The question can be rephrased as

The number of natural numbers which are less than 900 and relatively prime to it is

N=[1−(1a)][1−(1b)][1−(1c)]

N=[1−(12)][1−(13)][1−(15)]=240

### Question 7

Find the number of 4 digit numbers which are divisible by 11 or 13 but not by 17?

#### SOLUTION

Solution :B

Following are the steps while calculating the answer

4 digit numbers divisible by 11=899911=819, 4 digit numbers divisible by 13=899913=692 or 693

Number of integers common to 11 and 13 (as questionis 11 or 13) =8999(13×11)=62 or 63

To subtract numbers which are common to 11&17;13&17;11,13&17

11&17=8999(11×17)=47 or 48 terms

13&17=8999(13×17)=40 or 41

11,13,17=8999(11×13×17)=3 or 4 terms

Total ( you can approximate as the answers are far apart)= 819+693-63-48-41+4= 1364

Closest answer is option (2)

### Question 8

N=(323232.......50 digits)9. ie in base 9. Find the remainder when N is divided by 8?

#### SOLUTION

Solution :D

In base 9, divisibility check for 8 is sum of the digits(digit sum) =Digit sum of N=(25∗3+25∗2)9=

First convert the numbers to base 10

25 in base 10=23, Thus the question changes to (23∗3+23∗2)10=11510

Converting it back to base 9 = 137, Now sum of digits in base 9=1+3+7=(4+7)9=12⇒1+28∣R=3

### Question 9

10x when divided by 13 remainder = -1. If x is a natural number less than 100 how many solutions are possible for x ?

#### SOLUTION

Solution :D

The first power of 10 for which a division by 13 gives a remainder -1 is x=3. the next power of 10 will be 109 . hence it falls in an AP with first term as 3 and last term as 99 with a common difference of 6

number of terms =(99−3)6+1=966+1=16+1=17 terms.

### Question 10

How many two digit odd numbers are there with 10 factors?

#### SOLUTION

Solution :E

10 factors implies that the number can be of the form

1) a9 ( Least odd number will be 39)

2) ab4 (Least odd number will be 34∗5=405)

Therefore, as the least number is 405, there are no two digit numbers of the specified form. The answer is 0.

If a number N can be factorized as N=am∗bn(a, b are the prime factors)

Then number of factors= (m+1)(n+1)

### Question 11

x is the smallest number such that x2 is a perfect square and is x3 is a perfect cube. Then, the number of divisors of x is

#### SOLUTION

Solution :C

x2 is a perfect square ⇒x2=a2. x3 is a perfect cube ⇒x3=b3. x2×x3=a2×b3⇒x=√6a2b3=ab√6b.

For x to be a natural number; 6b has to be a perfect square ⇒b=6⇒x=23×34⇒ number of divisors =20

### Question 12

What is the value of m , if a3+3a2–am+4 when divided by a−2, gives a remainder m?

#### SOLUTION

Solution :B

We know that Number = Quotient × Divisor + Remainder

If Divisor (a-2) equals 0, then Number = Remainder. This is the property we are going to use

Make the divisor=0; i.e. put a=2, Then a3+3a2−am+4=m where a=2

Thus, 3m= 24 and m=8.

### Question 13

(11100)2=(1001)x. What is the value of x?

#### SOLUTION

Solution :C

(11100)2=0+0+22+23+24=28

(1001)x=x0+x3=1+x3

1+x3=28=27+1=33+1,x=3.

### Question 14

The number X4531Y, where X and Y are single-digit numbers, is divisible by 72. Then X - Y is equal to

#### SOLUTION

Solution :E

If a number is divisible by 72, it is divisible by 8 and 9. To be divisible by 8, the number formed by the last three digits should be divisible by 8. Therefore, 31Y should be divisible by 8⇒ Y = 2. To be divisible by 9, the sum of the digits of the number should be divisible by 9⇒ X = 3. Therefore, X-Y = 1

### Question 15

Find the least number which when divided by 9, 10 and 11 give remainders of 2,3 and 5 respectively.

#### SOLUTION

Solution :E

Use Chinese remainder theorem,

Let the number be N. N is of the form 9A+2 (i.e. N divided by 9 gives a remainder 2)= 10B+3 (i.e. N divided by 10 gives a remainder 3) =11C+5 (i.e. N divided by 11 gives a remainder of 5)

We will first find the number which is of the form 9A+2=10B+3.--------(1)

Here for B=8, A=9, which are the first integer solutions.

Substituting this in equation (1) we get the value 83, the first number which when divided by 9 gives a remainder 2 and which when divided by 10 gives a remainder 3.

The numbers will fall in an AP with the first digit as 83 and common difference 90 (9*10). The general form of the AP is 83+90k when k=0,1,2.....

Now to include the 3rd condition(i.e divisor 11 and remainder 5) we can write 83+90k=11C+5 -------(2)

Find the first integer value of K which satisfies the equation such that C is also an integer.

Here for K = 5, C = 48, which are the first integer solutions. Substituting this in equation

(2) we get the value 90 (5) + 83= 533, the first number which when divided by 9 gives a remainder 2, which when divided by 10 gives a remainder 3 and which when divided by 11 gives a remainder 5.

Shortcut:- You may be tempted to mark option (a) after a first glimpse at the options.

Since the question asks for the least number, you should be careful while answering.

You can still back calculate using answer option (a)

option (a) 1523 satisfies the condition. Hence 1523-990 = 533 has to be the first such

number. Answer is option (e)

Also note that, the first number has to be < 990.