## Number System Questions and Answers Part-5

1. The number of prime factors in the expressions 64 × 86 × 108 × 1210 is:
a) 80
b) 64
c) 72
d) 48

Explanation: 64 × 86 × 108 × 1210
= (2 × 3)4 × (23 )6 × (2 × 5)8 × (22 × 3 )10
= 24 × 34 × 218 × 28 × 58 × 220 × 310
= 250 × 314 × 58
The total prime factors,
= 50 + 14 + 8 [By adding maximum power of prime factors.]
= 72

2. x is five digit number. The digit in ten thousands place is 1. the number formed by its digits in units and ten places is divisible by 4. The sum of all the digits is divisible by 3. If 5 and 7 also divide x, then x will be.
a) 14020
b) 12060
c) 10020
d) 10080

Explanation: Let the digits of x be
x = abcde
x = 1bcde [Given ten thousands place is 1.]
Now we can check the options as given that the sum of the all digit is divisible by 3.
10080 is the only number given in the option which satisfies all the given conditions.

3. A man sells chocolates which are in the boxes. Only either full box or half a box of chocolates can be purchased from him. A customer comes and buys half the number of boxes which the seller had plus half box more. A second customer comes and purchases half the remaining number of boxes plus half a box. After this the seller is left with no chocolate boxes. How many chocolate boxes the seller had initially?
a) 2
b) 3
c) 4
d) 3.5

Explanation: The best way to go through the options
Let there are initially 3 boxes then,
1st customer gets = $$\frac{3}{2}$$ + $$\frac{1}{2}$$ = 2
Remaining boxes = 3 - 2 = 1
2nd customer = $$\frac{1}{2}$$ + $$\frac{1}{2}$$ = 1
Option 'b' is correct.

4. If x + y + z = 0, then x3 + y3 + z3 is equal to :
a) 0
b) 3xyz
c) $$\frac{{{\text{xy}} + {\text{yz}} + {\text{zx}}}}{{{\text{xyz}}}}$$
d) xyz(xy + yz + zx)

Explanation:

x + y + z = 0
Cubing both side,
(x + y + z)3 = 0
x3 + y3 + z3 - 3xyz = 0 [using formula]
x3 + y3 + z3 = 3xyz

5. To write all the page numbers of a book, exactly 136 times digit 1 has been used. Find the number of pages in the book.
a) 190
b) 195
c) 210
d) 220

Explanation: From 1- 99 digit 1 is used 20 times. And From 100 - 199, 1 is used 120 times
So, from 1 to 199, 1 is used,
20 + 120 = 140 times
We need 136. So leave 199, 198, 197 and 196
Required pages = 195

6. Given, N = 98765432109876543210 ..... up to 1000 digits, find the smallest natural number n such that N + n is divisible by 11.
a) 2
b) 3
c) 4
d) 5

Explanation: For a no. to be divisible by 11,
Sum(odd digit nos) - Sum(even digit nos) = 0 or divisible by 11
If we look at 9876543210, the difference we get is 5
i.e. [(9 + 7 + 5 + 3 + 1) - (8 + 6 + 4 + 2 + 0) = 5]
The series is up to 1000 digit,
That means, $$\frac{{1000}}{{10}}$$  = 100 time 5,
then the difference will be 5 × 100 = 500
In order for the difference to be divisible by 11, we need to add 5 and the no will become 505
505 is divisible by 11

7. The remainder when 40 + 41 + 42 + 43 + ........ + 440 is divided by 17 is:
a) 0
b) 16
c) 4
d) None of these

Explanation: Let S be the sum of the expression,
S = 40 + 41 + 42 + 43 + ........ + 440
S = (1 + 4 + 16 + 64) + 44 (1 = 4 + 16 + 64) + ...... + 436 + 440
Since, (1 + 4 + 16 + 64) = 85, is divisible by 17. Hence, except 440 remaining expression is divisible by 17.
$$\frac{{{4^{40}}}}{{17}} \to \frac{{{{\left( {{4^4}} \right)}^{10}}}}{{17}} \to \frac{{{1^{10}}}}{{17}}$$
Remainder = 1

8. The remainder when 30 + 31 + 32 + 33 + . . . . . . . + 3200 is divided by 13 is:
a) 0
b) 4
c) 3
d) 12

Explanation:
\eqalign{ & {\text{The}}\,{\text{given}}\,{\text{expression}}\,{\text{is}}\,{\text{in}}\,{\text{GP}}\,{\text{series}} \cr & S = {3^0} + {3^1} + {3^2} + {3^3} + ........ + {3^{200}} \cr & S = {\frac{{ {{3^0} \times \left( {{3^{201}} - 1} \right)} }}{{ {3 - 1} }}} \cr & S = \frac{{ {{3^{201}} - 1} }}{2} \cr & S = \frac{{ {{{\left( {{3^3}} \right)}^{67}} - {1^3}} }}{2} \cr & S = \frac{{ {{{27}^{67}} - {1^3}} }}{2} \cr}
Since, (An - Bn) is divisible by (A - B), So, (2767 - 13) is divisible by (27 - 1) = 26
Hence, Expression is also divisible by 13 as it is divisible by 26
Given expression is divisible by 13 so the remainder will be 0

9. The distance between the house of Rajan and Raman is 900 km and the house of former is at 100th milestone where as the house of Raman is at 1000th milestone. There are total 901 milestone at a regular interval of 1 km each. When you go to Raman's house from the house of Rajan which are on same highway, you will find that if the last digit (i.e. unit digit) of 3 digit number on every milestone is same as the first (i.e. hundreds digit) of the number on the next milestone is same, then these milestones must be red colour and rest will be of black. Total number of red colour milestone is:
a) 179
b) 90
c) 189
d) 100

a) $$\frac{2}{5}$$
b) $$\frac{1}{4}$$
c) $$\frac{1}{3}$$
d) $$\frac{1}{6}$$
Explanation: Since, he has covered twice the distance which yet he has to cover. It means he has covered $$\frac{2}{3}$$ of the whole journey and remaining journey is $$\frac{1}{3}$$ rd.