# Free Basic Numbers Practice Test - CAT

### Question 1

Find the remainder when 10100 is divided by 7

#### SOLUTION

Solution :A

Option (a)

Euler's number of 7 ( a prime number) is 7−1=6. make the numerator power the nearest

multiple of 6. The question changes to

10967∣R×1047∣R

part a will yield a remainder 1 and 1047∣R can be found using frequency method

1017∣R=3

1027∣R=2

1037∣R=67∣R=−1

1047∣R=47∣R=4

answer = option a

### Question 2

How many numbers are there from 1 to 1000 which when successively divided by 5, 6, 7 give remainders 2,4,5.

#### SOLUTION

Solution :C

option c

The first number in this AP can be found as given above

The first term = ((((5*6) +4)*5) + 2

Common difference = 5*6*7=210

General form of the AP is therefore 172+210k,

So the numbers will be 172, 382,592 and 802

### Question 3

Find the least number with 32 factors and at most 3 prime factors?

#### SOLUTION

Solution :C

option (c)

Let the prime factors be abc

Ways of writing

a31 least value =231

ab15 least value =2153

a3b7 least value =2733

ab3c3 least value =23335=1080

ab7c least value =273151=1920

### Question 4

A 32-digit number has all 9’s. Find the remainder when the number is divided by 111.

#### SOLUTION

Solution :D

option d

Any number of the form aaa (3 digits) or aaaaaa (multiple of 3 digits, i.e. 6, 9 .....) is divisible by 111

Therefore group the above numbers into groups of 3 or multiples of 3.

A 32 digit number can be broken up as (3*10 digits) + (2 digits). The first part is divisible by 111. the last two digits will thus be the remainder = 99

### Question 5

Find the remainder when (42)^{442}is divided by 100

#### SOLUTION

Solution :D

option (d)

Using the last two digits technique

(42)442=(21×2)442

=(21)442×(2)442=(21)442×(244)10×22=__41×__76×__04=__41×__04=__64. Option (d)

### Question 6

N=9876543298765432.........82 digits. Find the remainder when N is divided by 34?

#### SOLUTION

Solution :D

option d

Use Divisibility Rules

34 = 17 *2

So we should use the divisibility test of 17 ( Compartmentalization method - taking 8 digits at a time (Sum of digits at odd places taken 8 at a time- Sum of digits at even places taken 8 at a time)

Combine 8 digits together, 9876......80 digits*100 + 98 :

There will be equal number of groups (of 8 digits taken at a time) at odd places and even

Places in 9876.......80 digits*100.

So Sum of groups at odd places - Sum of groups at even places = 0

Therefore first part is divisible by 17. It is also divisible by 2, as 100 is divisible by 2.Hence, we only need to find the remainder when 98 is divided by 34

Remainder= 30

### Question 7

Find the remainder when 17^{36} is divided by 36?

#### SOLUTION

Solution :B

Answer=b

Approach 1

Go by frequency method

1736 gives a remainder =17

17236 gives a remainder =1

Frequency =2

Odd powers remainder =17, even powers =1

Approach 2

Euler’s number of 36 is 36×12×23=12

36=12k. hence, from Fermat theorem 173636|R=1

Note :

Euler's number is the number of co_primes of number which is less than that number.

A number N can be written as ambn (where a and b are the prime factors of N)

eg) 20=22×51

Here a=2 and b=5, m=2 and n=1

Euler's number =N[1−(1a)][1−(1b)]

From fermat Theorem

N(Euler′s number of y+k)y∣R=1 (i.e When a number N is raised to the Euler's number of a number)

"y" is divided by "y", the remainder will be 1

### Question 8

Find the last two digits of 7^{1 }+ 7^{2} + 7^{3} +……………..7^{342} ?

#### SOLUTION

Solution :E

option (e)

71=0775=−−07

72=4976=−−49

73=4377=−−43

74=0178=−−01

There is a cyclicity of 4 as 07, 49 , 43 ,01 is the last 2 digits for the powers of 7

Here 342=4k+2⇒Required answer is 07+49=56

### Question 9

What are the last two digits of 475^{23} ?

#### SOLUTION

Solution :C

option (c )

A number ending in 5, with the tens digit an odd number and raised to an odd power, will

always end with 75. Answer is option (c)

### Question 10

Solve for x

x=1+13+12+13+12+....∞

#### SOLUTION

Solution :D

1+13=43=1.333. the value of x will be slightly lesser than 1.333 as the denominator keeps increasing

Look at the options

Option a=51.7>2 .this can never be the answer

Option b=11.7<1. this also can never be the answer

Option c=31.4>2. this cannot be the answer

Option d=2.51.7 which is the correct answer

Option e=2, which can never be the answer