NUMBER SYSTEM NOTES
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No. of Questions 
20 
Time 
20 min 
Medium 
english 
Marks 
20 
Positive Marks 
+1 
Negative Marks 
0 
INSTRUCTION : All question carry 1 mark and there is only single option correct for every questionMark review to see the question later 
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Question 1 of 20
1. Question
1 pointsIf we write down all the natural numbers from 259 to 492 side by side get a very large natural number 259260261….491492259260261….491492.
How many 8’s will be used to write this large natural number?
Correct
Solution:
From 259 to 458, there are 200 natural numbers so there will be 2×20=402×20=40 8’s
From 459 to 492 we have 13 more 8’s and so answer is 40+13=40+13= 53
Incorrect
Solution:
From 259 to 458, there are 200 natural numbers so there will be 2×20=402×20=40 8’s
From 459 to 492 we have 13 more 8’s and so answer is 40+13=40+13= 53

Question 2 of 20
2. Question
1 pointsA girl wrote all the numbers from 100 to 200. Then she started counting the number of one’s that has been used while writing all these numbers. What is the number that she got?
Correct
Solution:
From 100 to 200 there are 101 numbers.
There are 100, 1’s in the hundred place.
10, 1’s in tens place and10, 1’s in unit place.Thus, the answer is 100+10+10=100+10+10= 120.
Incorrect
Solution:
From 100 to 200 there are 101 numbers.
There are 100, 1’s in the hundred place.
10, 1’s in tens place and10, 1’s in unit place.Thus, the answer is 100+10+10=100+10+10= 120.

Question 3 of 20
3. Question
1 pointsWhat is the value of M and N respectively if M39048458N is divisible by 8 and 11, where M and N are single digit integers?
Correct
Solution:
A number is divisible by 8 , if the number formed by the last three digits is divisible by 8.
i.e 58N is divisible by 8 ⇒ N=4Again a number is divisible by 11, if the difference between the sum of digits at even places and sum of digits at the odd places is either 0 or divisible by 11.
i.e, (M+9+4+4+8)−(3+0+8+5+N)=M−N+9=M+5(M+9+4+4+8)−(3+0+8+5+N)=M−N+9=M+5
It cannot be zero hence, M+5=11 ⇒ M=6.Incorrect
Solution:
A number is divisible by 8 , if the number formed by the last three digits is divisible by 8.
i.e 58N is divisible by 8 ⇒ N=4Again a number is divisible by 11, if the difference between the sum of digits at even places and sum of digits at the odd places is either 0 or divisible by 11.
i.e, (M+9+4+4+8)−(3+0+8+5+N)=M−N+9=M+5(M+9+4+4+8)−(3+0+8+5+N)=M−N+9=M+5
It cannot be zero hence, M+5=11 ⇒ M=6. 
Question 4 of 20
4. Question
1 pointsThe largest number amongst the following that will perfectly divide 101^{100} – 1 is:
Correct
Answer: Option ‘B’
The easiest way to solve such problems for objective exam purposes is trial and error or by back
substituting answers in the choices given.
101^{2} = 10,201
101^{2} − 1 = 10,200.
This is divisible by 100.
Similarly try for 101^{3} − 1 = 1,030,301−1 = 1,030,300.
So you can safely conclude that (101^{1} − 1) to (101^{9} − 1) will be divisible by 100.
(101^{10} − 1) to (101^{99} − 1) will be divisible by 1000.
Therefore, (101^{100} − 1) will be divisible by 10,000.
Incorrect
Answer: Option ‘B’
The easiest way to solve such problems for objective exam purposes is trial and error or by back
substituting answers in the choices given.
101^{2} = 10,201
101^{2} − 1 = 10,200.
This is divisible by 100.
Similarly try for 101^{3} − 1 = 1,030,301−1 = 1,030,300.
So you can safely conclude that (101^{1} − 1) to (101^{9} − 1) will be divisible by 100.
(101^{10} − 1) to (101^{99} − 1) will be divisible by 1000.
Therefore, (101^{100} − 1) will be divisible by 10,000.

Question 5 of 20
5. Question
1 pointsWhat is the unit place digit in (24)^{13} x (13)^{4} x (21)^{16}
Correct
Explanation:
In x^n, x is 4, if n is even the number is 6, n is odd then 4.
(24)^13 = units place in expansion is 4.Incorrect
Explanation:
In x^n, x is 4, if n is even the number is 6, n is odd then 4.
(24)^13 = units place in expansion is 4. 
Question 6 of 20
6. Question
1 pointsIn a division sum, the remainder is 6 and the divisor is 5 times the quotient and is obtained by adding 2 to the thrice of the remainder. The dividend is:
Correct
Divisor = (6 * 3) + 2 = 205 * Quotient = 20
Quotient = 4.
Dividend = (Divisor * Quotient) + Remainder
Dividend = (20 * 4) + 6 = 86.
Incorrect
Divisor = (6 * 3) + 2 = 205 * Quotient = 20
Quotient = 4.
Dividend = (Divisor * Quotient) + Remainder
Dividend = (20 * 4) + 6 = 86.

Question 7 of 20
7. Question
1 pointsIf n is a positive integer, which one of the following numbers must have a remainder of 3 when divided by any of the numbers 4, 5 and 6?
Correct
Solution:
Let m be a number that has a remainder of 3 when divided by any of the numbers 4, 5 and 6.
Then m−3m−3 must be exactly divisible by all three numbers.
Hence, m−3m−3 must be a multiple of the Least Common Multiple of the numbers 4, 5 and 6.
The LCM is 3×4×5=603×4×5=60.Hence, we can suppose m−3=60pm−3=60p, where pp is a positive integer.
Replacing pp with nn, we get m–3=60nm–3=60n.
So, m=60n+3m=60n+3.format 120n + 3 =60(2n)+3
Incorrect
Solution:
Let m be a number that has a remainder of 3 when divided by any of the numbers 4, 5 and 6.
Then m−3m−3 must be exactly divisible by all three numbers.
Hence, m−3m−3 must be a multiple of the Least Common Multiple of the numbers 4, 5 and 6.
The LCM is 3×4×5=603×4×5=60.Hence, we can suppose m−3=60pm−3=60p, where pp is a positive integer.
Replacing pp with nn, we get m–3=60nm–3=60n.
So, m=60n+3m=60n+3.format 120n + 3 =60(2n)+3

Question 8 of 20
8. Question
1 pointsA certain number when divided by 222 leaves a remainder 35, another number when divided by 407 leaves a remainder 47. What is the remainder when the sum of these two numbers is divided by 37?
Correct
Solution:
N1=222x+35N1=222x+35 and N2=407y+47N2=407y+47N1+N2=(37×6×x+35)+(37×11×y+47)N1+N2=(37×6×x+35)+(37×11×y+47)
Remainder when N1+N2N1+N2 is divided by 3737,
=35+4737=8
Incorrect
Solution:
N1=222x+35N1=222x+35 and N2=407y+47N2=407y+47N1+N2=(37×6×x+35)+(37×11×y+47)N1+N2=(37×6×x+35)+(37×11×y+47)
Remainder when N1+N2N1+N2 is divided by 3737,
=35+4737=8

Question 9 of 20
9. Question
1 pointsThe sum of three digit number is subtracted from the number. The resulting number is always:
Correct
Solution:
Let the three digit number be 439Sum of digits =16=16
Difference =439−16=423which is divisible by 9.Incorrect
Solution:
Let the three digit number be 439Sum of digits =16=16
Difference =439−16=423which is divisible by 9. 
Question 10 of 20
10. Question
1 pointsWhat is the last digit of the number 3^{579}+ 1?
Correct
Solution:
Any power of 5 when divided by 4 gives a remainder 1.Here, the power of 3 is itself a power of 5 and will give the remainder of 1 when divided by 4.
The last digit of the number will be 3.
And, hence, the last digit of the given number is 3+1=3+1= 4.
Incorrect
Solution:
Any power of 5 when divided by 4 gives a remainder 1.Here, the power of 3 is itself a power of 5 and will give the remainder of 1 when divided by 4.
The last digit of the number will be 3.
And, hence, the last digit of the given number is 3+1=3+1= 4.

Question 11 of 20
11. Question
1 pointsTwo prime numbers A,B(A<B) are called twin primes if they differ by 2 (e.g. 11,13,or 41,43….). If A and B are twin primes with B>23, then which of the following numbers would always divide A+B?
Correct
Solution:
Any prime number greater than 3 will be in the form of 6x+16x+1 or 6x−16x−1.Thus, both prime number are twins:
Let first be 6x−1
and 2nd be 6x+1
Sum=12xThus it is always divisible by 12.
Incorrect
Solution:
Any prime number greater than 3 will be in the form of 6x+16x+1 or 6x−16x−1.Thus, both prime number are twins:
Let first be 6x−1
and 2nd be 6x+1
Sum=12xThus it is always divisible by 12.

Question 12 of 20
12. Question
1 pointsWhat is the remainder left after dividing 1!+2!+3!.....+100! by 77?
Correct
Solution:
7!+8!+9!.......+100! is completely divisible by 77.
Now, 1!+2!+3!....+6!=873
When 873 is divided by 77 it leaves a remainder =5.
Incorrect
Solution:
7!+8!+9!.......+100! is completely divisible by 77.
Now, 1!+2!+3!....+6!=873
When 873 is divided by 77 it leaves a remainder =5.

Question 13 of 20
13. Question
1 pointsWhen a particular positive number is divided by 5, the remainder is 2. If the same number is divided by 6, the remainder is 1. If the difference between the quotients of division is 3, then find the number.
Correct
Solution:
Let the quotients when this number is divided by 5 and 6 be x and y respectively.
(Note that x will be greater than y as 5 is smaller than 6).
Number =5x+2=6y+1
Given that,x−y=3
On solving both equation we get, x=19,y=16
Thus the number is 19×5+2= 97.
Incorrect
Solution:
Let the quotients when this number is divided by 5 and 6 be x and y respectively.
(Note that x will be greater than y as 5 is smaller than 6).
Number =5x+2=6y+1
Given that,x−y=3
On solving both equation we get, x=19,y=16
Thus the number is 19×5+2= 97.

Question 14 of 20
14. Question
1 pointsThe numbers 1 to 29 are written side by side as follows 1234567891011……2829. If the number is divided by 9, then what is the remainder?
Correct
Solution:
Sum of digits from 11 to 10=46
sum of digits from 1111 to 20=56
sum of digits from 2121 to 29=63
sum of the digits from numbers =46+56+63=165
Sum of digits in the number 165=12 which gives a remainder of 3 when divided by 99.
Incorrect
Solution:
Sum of digits from 11 to 10=46
sum of digits from 1111 to 20=56
sum of digits from 2121 to 29=63
sum of the digits from numbers =46+56+63=165
Sum of digits in the number 165=12 which gives a remainder of 3 when divided by 99.

Question 15 of 20
15. Question
1 pointsHow many zeroes will be there in the expansion of the expression
(1)^1 x ( 2)^2 x (3)^3………(100)^100
Correct
Solution:
Number of zeroes will be decided by the power of 2 and 5 in the product.
Since, the power of 5 is less than the power of 2 hence number of zero will be equal to power of 5.
Power of 5 =
⇒ (5+10+15+20+(25×2)+30+40+.....)
⇒ (5+10+15+....100)+(25+50+75+100)⇒ 202[2×5+19×5]+250
⇒ 1050+250= 1300Incorrect
Solution:
Number of zeroes will be decided by the power of 2 and 5 in the product.
Since, the power of 5 is less than the power of 2 hence number of zero will be equal to power of 5.
Power of 5 =
⇒ (5+10+15+20+(25×2)+30+40+.....)
⇒ (5+10+15+....100)+(25+50+75+100)⇒ 202[2×5+19×5]+250
⇒ 1050+250= 1300 
Question 16 of 20
16. Question
1 pointsA number when divided by a divisor leaves a remainder of 24. When twice the original number is divided by the same divisor, the remainder is 11. What is the value of the divisor?
Correct
Solution:
Let the original number be ‘a‘
Let the divisor be ‘d‘Let the quotient of the division of a by d be ‘x‘
Therefore, we can write the relation as a/d=x and the remainder is 24.
i.e., a=dx+24When twice the original number is divided by d, 2a is divided by d.
We know that a=dx+24. Therefore, 2a=2dx+48The problem states that (2dx+48)/d leaves a remainder of 11.
2dx is perfectly divisible by dd and will therefore, not leave a remainder.
The remainder of 11 was obtained by dividing 48 by dd.
When 48 is divided by 37, the remainder that one will obtain is 11.
Hence, the divisor is 37.
Incorrect
Solution:
Let the original number be ‘a‘
Let the divisor be ‘d‘Let the quotient of the division of a by d be ‘x‘
Therefore, we can write the relation as a/d=x and the remainder is 24.
i.e., a=dx+24When twice the original number is divided by d, 2a is divided by d.
We know that a=dx+24. Therefore, 2a=2dx+48The problem states that (2dx+48)/d leaves a remainder of 11.
2dx is perfectly divisible by dd and will therefore, not leave a remainder.
The remainder of 11 was obtained by dividing 48 by dd.
When 48 is divided by 37, the remainder that one will obtain is 11.
Hence, the divisor is 37.

Question 17 of 20
17. Question
1 pointsLet n be the number of different 5 digit numbers, divisible by 4 with the digits 1, 2, 3, 4, 5 and 6, no digit being repeated in the numbers. What is the value of n?
Correct
Solution:
Test of divisibility by 4 is that the last two digits should be divisible by 4.
Case 1 : When the last 2 digits are 12, i.e., _ _ _12=4×3×2=24 numbers
Case 2 : When the last 2 digits are 16, there are 24 numbers
Case 3 : When the last 2 digits are 24 there are 24 numbers
Case 4 : When the last 2 digits are 32 there are 24numbers
Case 5 : When last 2 digits are 36 there are 24 numbers
Case 6 : When last 2 digits are 52 there are 24 numbers
Case 7 : When last 2 digits are 56 there are 24 numbers
Case 8 : When last 2 digits are 64 there are 24 numbers
Total =8×24= 192
Incorrect
Solution:
Test of divisibility by 4 is that the last two digits should be divisible by 4.
Case 1 : When the last 2 digits are 12, i.e., _ _ _12=4×3×2=24 numbers
Case 2 : When the last 2 digits are 16, there are 24 numbers
Case 3 : When the last 2 digits are 24 there are 24 numbers
Case 4 : When the last 2 digits are 32 there are 24numbers
Case 5 : When last 2 digits are 36 there are 24 numbers
Case 6 : When last 2 digits are 52 there are 24 numbers
Case 7 : When last 2 digits are 56 there are 24 numbers
Case 8 : When last 2 digits are 64 there are 24 numbers
Total =8×24= 192

Question 18 of 20
18. Question
1 pointsThree gold coins of weight 780gm, 840gm and 960gm are cut into small pieces, all of which have the equal weight. Each piece must be heavy as possible. If one such piece is shared by two persons, then how many persons are needed to give all the pieces of gold coins?
Correct
Solution:
HCF(780,840,960)=60Thus, total number of pieces:
⇒78060+84060+96060=13+14+16=43Total number of person required =43×2= 86.
Incorrect
Solution:
HCF(780,840,960)=60Thus, total number of pieces:
⇒78060+84060+96060=13+14+16=43Total number of person required =43×2= 86.

Question 19 of 20
19. Question
1 pointsWhat is the unit digit in {(6374)^{1793} x (625)^{317} x (341^{491})}?
Correct
Solution:
Unit digit in (6374)^{1793} = Unit digit in (4)^{1793}
= Unit digit in [(4^{2})^{896} x 4]
= Unit digit in (6 x 4) = 4
Unit digit in (625)^{317} = Unit digit in (5)^{317} = 5
Unit digit in (341)^{491} = Unit digit in (1)^{491} = 1
Required digit = Unit digit in (4 x 5 x 1) = 0.
Incorrect
Solution:
Unit digit in (6374)^{1793} = Unit digit in (4)^{1793}
= Unit digit in [(4^{2})^{896} x 4]
= Unit digit in (6 x 4) = 4
Unit digit in (625)^{317} = Unit digit in (5)^{317} = 5
Unit digit in (341)^{491} = Unit digit in (1)^{491} = 1
Required digit = Unit digit in (4 x 5 x 1) = 0.

Question 20 of 20
20. Question
1 pointsSix bells commence tolling together and toll at intervals of 2, 4, 6, 8 10 and 12 seconds respectively. In 30 minutes, how many times do they toll together ?
Correct
Explanation:
L.C.M. of 2, 4, 6, 8, 10, 12 is 120.
So, the bells will toll together after every 120 seconds(2 minutes).
In 30 minutes, they will toll together(30/2)+ 1 = 16 times.
Incorrect
Explanation:
L.C.M. of 2, 4, 6, 8, 10, 12 is 120.
So, the bells will toll together after every 120 seconds(2 minutes).
In 30 minutes, they will toll together(30/2)+ 1 = 16 times.