Exams > Cat > Quantitaitve Aptitude
BASIC NUMBERS MCQs
Total Questions : 20
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Answer: Option C. -> 1080
:
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
:
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
Answer: Option D. -> 64
:
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)
:
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)
Answer: Option C. -> 75
:
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)
:
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)
Answer: Option A. -> 4
:
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
:
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
Answer: Option D. -> 30
:
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
:
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
Answer: Option E. -> 56
:
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
:
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
Answer: Option B. -> 1
:
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′snumberofy+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
:
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′snumberofy+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
Answer: Option D. -> 99
:
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
:
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
Answer: Option D. -> 2517
:
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
:
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