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BASIC NUMBERS MCQs

Total Questions : 20 | Page 1 of 2 pages
Question 1. Find the least number with 32 factors and at most 3 prime factors?
  1.    1920
  2.    2733
  3.    1080
  4.    582
  5.    none of these
 Discuss Question
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
Question 2. How many numbers are there from 1 to 1000 which when successively divided by 5, 6, 7 give remainders 2,4,5.
  1.    1
  2.    3
  3.    4
  4.    6
  5.    5
 Discuss Question
Answer: Option C. -> 4
:
C
option c
The first number in this AP can be found as given above
How Many Numbers Are There From 1 To 1000 Which When Success...
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 remainder when (42)442is divided by 100
  1.    24
  2.    34
  3.    44
  4.    64
  5.    74
 Discuss Question
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)
Question 4. What are the last two digits of 47523 ?
  1.    25
  2.    50
  3.    75
  4.    85
  5.    None of these
 Discuss Question
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)
Question 5. Find the remainder when 10100 is divided by 7
  1.    4
  2.    3
  3.    2
  4.    -1
  5.    none of these
 Discuss Question
Answer: Option A. -> 4
:
A
Option (a)
Euler's number of 7 ( a prime number) is 71=6. make the numerator power the nearest
multiple of 6. The question changes to
10967R×1047R
part a will yield a remainder 1 and 1047R can be found using frequency method
1017R=3
1027R=2
1037R=67R=1
1047R=47R=4
answer = option a
Question 6. N=9876543298765432.........82 digits. Find the remainder when N is divided by 34?
  1.    0
  2.    1
  3.    15
  4.    30
  5.    -1
 Discuss Question
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
Question 7. Find the last two digits of 71 + 72 + 73 +……………..7342 ?
  1.    07
  2.    01
  3.    49
  4.    43
  5.    56
 Discuss Question
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+2Required answer is 07+49=56
Question 8. Find the remainder when 1736 is divided by 36?
  1.    0
  2.    1
  3.    -1
  4.    17
  5.    2
 Discuss Question
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(Eulersnumberofy+k)yR=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 9. A 32-digit number has all 9’s. Find the remainder when the number is divided by 111.
  1.    9
  2.    11
  3.    1
  4.    99
  5.    none of these
 Discuss Question
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
Question 10. Solve for x
x=1+13+12+13+12+....
  1.    5017
  2.    1017
  3.    3014
  4.    2517
  5.    2
 Discuss Question
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

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