# Digits and Numbers

We will discuss here how to solve different types of problems related to digits and numbers.

1. The number of digits in (48^4 × 5^12) is

(a) 18

(b) 16

(c) 14

(d) 12

Solution:

48^4 × 5^12

= (16 × 3)^4 × 5^12

= (2^4)^4 × (3)^4 × (5)^12

= (2)^16 × (3)^4 × (5)^12

= (2 × 5)^12 × (2 × 3)^4

= (10)^12 × (6)^4

= 1296 × 10^12

Therefore, the required number of digits = 4 + 12 = 16

2. All the prime numbers from 2 to 200 are multiplied together. How many zeros are there at the end of the product on the right?

(a) 21

(b) 22

(c) 24

(d) 25

Solution:

Number of zeros at the end of 2 × 4 × 6 × 8 × ........... × 200

= 200/(5 × 2) + 200/(5^2 × 2)

= 20 + 4

= 24

Note: Number of zeros at the end of the product 2 × 4 × 6 × 8 × ........... 2n = 2n/(5 × 2) + 2n/(5^2 × 2) + 2n/(5^3 × 2) + ...............

3. Numbers 10, 20, 30, 40, ........ , 980, 990, 1000 are multiplied together. The number of zeros at the end of the product (on the right) will be

(a) 124

(b) 120

(c) 111

(d) 110

Solution:

Number of zeros at the end of 10 × 20 × 30 × 40 × ............ × 1000 = 1000/(5 × 2) + 1000/(5^2 × 2) + 1000/(5^3 × 2) = 100 + 20 + 4 = 124

Note: Number of zeros at the end of the product of 10 × 20 × 30 × 40 × ............ × 10n = 10n/(5 × 2) + 10n/(5^2 × 2) + 10n/(5^3 × 2) + .........

4. The quotient of two positive integers is 9/5 and their product is 11520. The difference of these two numbers is:

(a) 60

(b) 64

(c) 74

(d) 70

Solution:

Quotient of division = 9/5

Therefore, the ratio of two numbers = 9 : 5

Now, 9x + 5x = 11520

or, 45x^2 = 11520

or, x^2 = 256

or, x = 16

Therefore, the required difference (9x - 5x) 4x = 4 × 16 = 64

5. The sum of a rational number and the reciprocal of that rational number is 13/6. The number is:

(a) 12/13

(b) 5/6

(c) 3/2

(d) 13/12

Solution:

Let, the rational number be a/b

Therefore, its reciprocal number be b/a

Now, a/b + b/a = 13/6

or, (a^2 + b^2)/ab = 13/6

or, (a^2 + b^2)/ab = (3^2 + 2^2)/(3 × 2)

Therefore, the required number = 3/2

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