number systems problems with explanations - gr8ambitionz

ww

w.G

r8A

mbi

tionZ

.com

for more materials and practice sets visit us at www.Gr8AmbitionZ.com

1 Number Systems Practice Problems with explanations – powered by www.Gr8AmbitionZ.com -

ww

w.G

r8A

mb

itio

nZ

.co

m

PROBLEMS ON NUMBER SYSTEMS – with explanations

These questions will prove to be good exercise for some good practice in number systems. Try to do the questions without looking at the solutions. The time allotted is 30 mins. All The Best Question 1

The sum of , , , … is : (1) less than 1 (2) greater than 1

(3) equal to 1 or less than 1 (4) equal to 1/n2 (n + 2)

The right answer is Option (1). ¼ + 1/9 + 1/16 +…+ 1/n2

Consider ¼:

¼ = 1/(2 x 2) < 1/(1 x 2) because ½ < 1

Similarly, 1/9 = 1/(3 x 3) < 1/(2 x 3)

Since 1/3 < ½

1/16 = 1/(4 x 4) < 1/(3 x 4)

Because ¼ < 1/3

……………………………………………………..

……………………………………………………..

1/n2 = 1/(nxn) < 1/(n – 1)n

Since 1/n < 1/(n – 1)

¼ + 1/9 + …+ 1/n2 < 1/(1 x 2) + 1/(2 x 3) + … + 1/[(n – 1)n]

i.e. ¼ + 1/9 + 1/16 + … + 1/n2 < (1/1 – ½) + (1/2 – 1/3) +…+ [1/(n – 1) – 1/n]

i.e. ¼ + 1/9 + 1/16 +… + 1/n2 < 1/1 – 1/n which is less than 1.

Hence, the given sum is less than 1 only.

Answer: (1)

ww

w.G

r8A

mbi

tionZ

.com



ww

w.G

r8A

mb

itio

nZ

.co

m

Question 2 What is the digit in the unit place of 3946252 + 734733 ?

(1) 8 (2) 7

(3) 6 (4) 4

The right answer is Option (1). 3946252 + 734733

First we will find the unit digit of 3946252.

Divide 46, by 4, we get 2 as remainder.

Actual remainder (2)252

Again dividing (2)252 by 4, we get 0 as remainder. And ‘9’ is an odd digit number. So, the last digit

of 3946252 will be 1.

Last digit of 734733 By dividing 47 by 4, we get 3 or – 1 as a remainder.

Now, (– 1) odd, when divided by 4 will give – 1 or 3 as remainder.

So, the last digit of 734733 will be.

Last digit of 3 3 3 = i.e. 7.

So, the last digit is 1 + 7 = 8. Answer: (1)

Question 3

How many terms, at the minimum of sequence 1, , , , …., must be added together for their sum to be not less than 3? (1) 16 (2) 17

(3) 18 (4) 20

The right answer is Option (1). 1 + ½ + 1/3 + ¼ + 1/5 + 1/6 + 1/7 + …

Grouping the terms as follows

(1) + (1/2) + (1/3 + ¼) + (1/5 + 1/6 + 1/7 + 1/8) + (1/9 + 1/10 + … 1/16) + …

(1/3 + ¼) > 2 x ¼, i.e. (1/3 + ¼) > ½

Similarly (1/5 + 1/6 + 1/7 + 1/8) > 4 x 1/8, i.e. > ½

Similarly (1/9 + 1/10 + … + 1/16) > 8 x 1/16,

i.e. > ½

If you take 16 terms their sum will be greater than 3. Answer: (1)

Question 4 When we divide 333435 by 34 the remainder is a and when we divide 333534 by 34 the remainder is b. What is the value of b – a? (1) 0 (2) 32

(3) -32 (4) None of these

ww

w.G

r8A

mbi

tionZ

.com



ww

w.G

r8A

mb

itio

nZ

.co

m

The right answer is Option (2).

333435 = (34 – 1)some even power =

333534 = (34 – 1)some odd power =

Now in first case, remainder a = 1

In second case, remainder b = 33

b – a = 33 – 1 = 32. Answer: (2)

Question 5 ab and cd are two two-digit natural numbers. 4b + a = 13k1 and 5d – c = 17k2, where k1 and k2 are natural numbers. The largest number that will always divide the product of ab and cd is: (1) 13 (2) 17

(3) 221 (4) None of these

The right answer is Option (3). ab is always a multiple of 13 and cd is always a multiple of 17 so their product has to be divisible

by

17 13 = 221. Answer: (3)

Question 6 What is the number, the sum of whose factors is equal to

? (1) 1124695 (2) 1123685

(3) 1012675 (4) None of these

The right answer is Option (1). The sum of all the factor is given by

=

=

it means

So, the number is 132 113 51

169 1331 5 = 1124695. Answer: (1)

ww

w.G

r8A

mbi

tionZ

.com



ww

w.G

r8A

mb

itio

nZ

.co

m

Question 7

A is the smallest integer which when multiplied with 3 gives a number made of 5’s only. Sum of the digits of A is B. Sum of the digits of B is C. What is the value of C3? (1) 125 (2) 64



Smallest number made of 5’s and divisible by 3 is 555. Thus A = = 185. Answer: (1)

Question 8 3762 – 2362 is divisible by:

(1) 560 (2) 840

(3) 775 (4) 160

The right answer is Option (2). 3762 – 2362

xn – yn is always divisible by (x – y)

xn – yn is divisible by (x + y) when n is even.

So, 3762 – 2362 will be divisible by both (37 – 23) and (37 + 23)

So, 3762 – 2362 will be divisible by (37 – 23) x (37 + 23) = 14 60 = 840

Answer: (2)

Question 9

Find the smallest value of x such that x! ends with exactly 23 zeroes

(1) 100 (2) 98

(3) 99 (4) No such no. exist

The right answer is Option (4). If we check that, 99! ends with how many zeroes, then we find that it ends with 22 zeroes

19 + 3 = 22

So, Now if we find that 100! ends with how many zeroes then it comes out be 24.

20 + 4 = 24

So, there are no such no. exists with ends with 23 zeroes. Answer: (4)

ww

w.G

r8A

mbi

tionZ

.com



ww

w.G

r8A

mb

itio

nZ

.co

m

Question 10 abc is a three-digit whole number so that abc = a3 + b3 + c3. [300 < abc < 400]. Find the sum of a, b & c. (1) 10 (2) 11

(3) 12 (4) Data insufficient

The right answer is Option (4). Since 300 < abc < 400

100a + 10b + c = a3 + b3 + c3

put a = 3 and rearrange, we get

300 + 10b + c = 27 + b3 + c3

(b3 – 10b) + (c3 – c) = 273

Now put b = 6 and 7 (as b and c are both less than equal to 9 and 63 and 73 are the nearest

cubes to 273).

For b = 6. We get c3 – c = 117 and no value of c satisfies this.

For b = 7 we get c3 – c = 0 and two values of c i.e. 0 and 1.

Thus abc = 370 or 371

Hence, abc cannot be uniquely determined.

Answer: (4)

Question 11 Two different numbers when divided by the same divisor, left remainder 11 and 21 respectively, and when their sum was divided by the same divisor, then the remainder was 4. Find the divisor. (1) 36 (2) 28

(3) 14 (4) 9

The right answer is Option (2). Let the numbers are x1, x2 when divided by y (divisor), quotients are n & m and remainders are 11

& 21

respectively. Therefore

x1 = ny + 11 …..(1)

x2 = my + 21 …..(2)

Adding (1) and (2), x1 + x2 = (n + m)y + 32 = (n + m)y + 28 + 4

As the remainder given is 4, this is possible if y divides 28.

So y can be 14 or 28 but 14 is not possible because in (2) the remainder is 21. Answer : (2)

ww

w.G

r8A

mbi

tionZ

.com



ww

w.G

r8A

mb

itio

nZ

.co

m

Question 12 Which of the following numbers is divisible by 23?

(1) 22! + 1 (2) 22! – 1

(3) 21! + 2 (4) None of these

The right answer is Option (1). According to the Wilson’s theorem

(P – 1)! + 1 is always divisible by P

Where P is a prime number.

So, we can say that

(23 – 1)! + 1 is divisible by 23

22! + 1 is divisible by 23. Answer: (1)

Question 13

Let N = 103 + 104 + 105 + 106 + 107 + 108 + 109. The sum of the digits of N is:

(1) 12 (2) 1

(3) 6 (4) 7

The right answer is Option (4). Since N is written out as a sum of powers of 10, then N can be written as 1111 111 000, and so

the sum of the digits is 7. Answer: (4)

Question 14

If 27 = 123 and 31 = 133, then 15 = ?

(1) 13 (2) 31

(3) 11 (4) 33

The right answer is Option (4). Let the base be N.

then 27 = (123)N

27 = N2 + 2 N1 + 3 No = N2 + 2N + 3

N2 + 2N – 24 = (N + 6) (N – 4) = 0 N = – 6, 4

But the base of a number system can’t be negative. Thus N = 4.

Now, 15 in base 4 can be written as (33)4. Answer : (4)

ww

w.G

r8A

mbi

tionZ

.com



ww

w.G

r8A

mb

itio

nZ

.co

m

Question 15 The expression 333555 + 555333 is divisible by:

(1) 2 (2) 3

(3) 37 (4) All of these

The right answer is Option (4). 333555 + 555333

Check it by option.

The last digit of 333555 will be odd, and the last digit of 555333 will be odd.

odd + odd even

So, the last digit is even, which is divisible by 2.

Both the numbers are individually divisible by 3. So 333555 + 555333 is divisible by 3.

And similarly, both the numbers are divisible by 37, so, 333555 + 555333 is divisible by 37.

Hence this expression is divisible by all three numbers. Answer : (4) Note : If the number consists of same in the multiples of 3, then it is always a multiple of 37

Question 16

The value of (1.00008)12500 is

(1) < 1 (2) > 1

(3) > 2 (4) > 3


(1.00008)12500 = (1 + .00008)12500 =

Applying binomial theorem: (a + b)n = nC0 an + nC1 an – 1 . b1 +……… + nCn – 1 . a1 . bn – 1 + nCn .

bn

= 1 + 12500 +…………….+

= 2 + …………….+ > 2. Answer : (3)

Question 17

The last two digits of 2563 6325 are :

(1) 35 (2) 75

(3) 55 (4) 45

The right answer is Option (2). Last two digits of 2563 are always 25.

ww

w.G

r8A

mbi

tionZ

.com



ww

w.G

r8A

mb

itio

nZ

.co

m

Last two digits of 6325 = an even no 3

( Ten’s digit of any power of a number, whose unit digit is odd and Ten’s digit is even, is always

even).

Thus, last two digits of 2563 6325 = _ _ _ 2 5 an even no 3 = 75. Answer : (2)

Question 18

The remainder when 1010 + 10100 + 101000 + … + 1010000000000 is divided by 7 is:

(1) 0 (2) 1

(3) 2 (4) 5

The right answer is Option (4). There are total of 10 terms.

Now apply congruence modulo theorem.

10 3 (mod 7)

10 – 4 (mod 7)

1010 = (–4)10 (mod 7)

= 410 (mod 7)

Now series can be re-written as.

410 + 4100 + 41000 + …………. + 410000000000

(43)3 4 + (43)33 4 + ……. + (43)333333333 4

Now 43 1 (mod 7)

Thus, remainder when above series is divided by 7 is

[(1) 4 + (1) 4 + ……. 10 times] = 40

40 5 (mod 7). Answer : (4)

Question 19

The number 6n2 + 6n for natural n is always divisible by

(1) 6 only (2) 18

(3) 12 only (4) 6 and 12

The right answer is Option (4). Since the above expression contains a 6, it is always divisible by 6. Also, n (n + 1) is divisible by

2.

Hence, 6n (n + 1) is divisible by 6 2 = 12. The correct answer is 12 and 6. Answer: (4)

Question 20

The number is (1) a finite number (2) an infinite non–repeating decimal

(3) equal to 1.7320508 (4) an infinite repeating decimal


ww

w.G

r8A

mbi

tionZ

.com



ww

w.G

r8A

mb

itio

nZ

.co

m

An infinite non–repeating decimal. Direct from definition. Answer: (2)

Question 21 Let D be a decimal of the form D = 0. abcd abcd abcd…………, where digits a, b, c and d are integers lying between 0 and 9. At most three of these digits are zero. By what number D be multiplied so that the result is a natural number? (1) 999 (2) 9990



D = ,

Also 49995 = 9999 5

Hence, once we multiply D with 49995 we will get a natural number. Answer: (3)

Question 22

N = (11111111)2. What is the sum of the digits of N?

(1) 72 (2) 62


The right answer is Option (3). 12 = 1, 112 = 121, 1112 = 12321

111111112 = 123456787654321. The sum of these numbers is 64. Answer: (3)

Question 23

What is the remainder when 192319241925 is divided by 1924?

(1) 1922 (2) 1923

(3) 1 (4) 1922

The right answer is Option (3). 192319241925 = (1924-1)some evenpower = x

Remainder = (–1)x = 1. Answer: (3)

Question 24

N = 2 4 6 8 10 ……. 100. How many zeroes are there at the end of N?

(1) 24 (2) 13


The right answer is Option (3). N = 250 50!. The highest power of 5 in 50! is 12, therefore there will be 12 zeroes at the end of

N. Answer: (3)

ww

w.G

r8A

mbi

tionZ

.com



ww

w.G

r8A

mb

itio

nZ

.co

m

Question 25 If in a number system, 40132 corresponds to 2542 in decimal, find the base of the number system. (1) 5 (2) 6

(3) 7 (4) 8

The right answer is Option (1). Since, four is one of the characters in the given number, the base for that number system must

be > 5.

Assume that the base = 5. Then, (4 54) + (0 53) + (1 52) + (3 51) + (2 50) = 2542.

So, by estimation, we can say that the base of the number system is 5. Answer: (1)

We are at Facebook

Don’t forget to LIKE us there for Daily Updates

www.facebook.com/Gr8AmbitionZ

number systems problems with explanations - gr8ambitionz

Documents