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
Answer: (b)
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
Answer: (c)
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
Answer: (a)
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
Answer: (b)
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
Answer: (c)
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