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CAT 2025 Lesson : Number Systems - Special Types

5.4 Skipping digits

Example 11

The micronanometer in a certain factory can measure the pressure inside the gas chamber from

1

unit to

999999

units. Lately this instrument has not been working properly. The problem with the instrument is that it always skips the digit

5

and moves directly from

4

6

. What is the actual pressure inside the gas chamber if the micronanometer displays

003016

?
[XAT 2011]
(1)

2201

(2)

2202

(3)

2600

(4)

2960

(5) None of the above

Solution

As the instrument always skips

5

, it uses

9

digits only. This means, the instrument uses a Base

9

number system.

Digits are nothing but symbols that are used to represent certain values. In this chapter we have learnt conversions or reconversions to Base

9

number system, where the

9

digits are

0, 1, 2, 3, 4, 5, 6, 7, 8.

The instrument, however, uses

0, 1, 2, 3, 4, 6, 7, 8, 9

. The nine digits in sequential order of the Base

9

system and that of the instrument are provided below.

Machine	$0$	$1$	$2$	$3$	$4$	$6$	$7$	$8$	$9$
Base $9$	$0$	$1$	$2$	$3$	$4$	$5$	$6$	$7$	$8$

Where the digits in the instrument is between

6

and

9

, if we reduce by

1

, we effectively convert the instrument number to a base

9

number (the way we've learnt base

9

).

In

3016

, we need to replace

6

with

5

.

∴

3016

of instrument

= (3015)_{9}

(3015)_{9} = (3 \times 729) + (1 \times 9) + 5 = 2201

Answer: (1)

2201

5.5 Special Type - Bases 2 and 3

Base 2 has only 2 digits - 0 and 1. This means a number written in base 2 is the sum of exponents of 2.

For instance,

(27)_{10}

(11011)_{2}

2^{4} + 2^{3} + 2^{1}+ 2^{0}

Therefore, if we have the numbers

a_{1} = 2^{0}, a_{2} = 2^{1}, a_{3} = 2^{2}, ... , a_{n} = 2^{n-1}

, then any integer between 1 and (

2^{n}

- 1), both inclusive, can be formed by adding one or more from

a_1

a_{n}

.

In fact,

a_{1} = 2^{0}

a_{n} = 2^{n-1}

results in the lowest value that n can take in order to be able form any integer between 1 and

(2^{n} - 1)

, both inclusive.

Therefore, if

n

is the number of digits of the upper limit of the range when written in base

2

, then

n

terms are required to form any integers between

1

and the upper limit.

Example 12

Ashish is given Rs. 158 in one rupee denominations. He has been asked to allocate them into a number of bags such that any amount required between Re. 1 and Rs. 158 can be given by handing out a certain number of bags without opening them. What is the minimum number of bags required?
[CAT 2001]

(1) 11 (2) 12 (3) 13 (4) None of these

Solution

2^{0}, 2^{1}, ... , 2^{6}

can be used to form numbers from 1 to 127 (i.e.,

2^{7}

- 1).

2^{0}, 2^{1}, ... , 2^{7}

can be used to form numbers from 1 to 255 (i.e.,

2^8

- 1). Integers between 1 and 158 can be formed using these.

So, Ashish can put 1, 2, 4, 8, 16, 32, 64 into 7 bags and the remaining 31 coins into the

8^{\mathrm{th}}

bag. Therefore, the minimum number of bags that are required is 8 bags.

Answer: (4) None of these

Now if you need to find n integers such that by adding or subtracting them you should be able to form numbers, then the integers can be used in three ways - adding it, subtracting it or not using it. As there are three possibilities, this is akin to a base 3 system, where the digits are 1, 0 and -1.

From the perspective of solving these questions all you need to note is the following. Using the integers

3^{0}

= 1,

3^{1}

= 3 and

3^{2}

= 9, we will be able to form all numbers from 1 to 13 (i.e., 1 + 3 + 9), as shown below.

1 = 1	2 = 3 - 1	3 = 3	4 = 3 + 1	5 = 9 - 3 - 1	6 = 9 - 3	7 = 9 - 3 + 1
8 = 9 - 1	9 = 9	10 = 9 + 1	11 = 9 + 3 - 1	12 = 9 + 3	13 = 9 + 3 + 1

Therefore, the fewest number of items we would be using is

3^{0}, 3^{1}, 3^{2}, ... , 3^{n - 1}

to form any number between 1 and sum of all these terms from

3^{0}

3^{n - 1}

, which is

\dfrac{3^n - 1}{2}

Example 13

Ram has a pan balance with which he should be able to measure all possible integral values in kilograms (kg) between 1 kg and 100 kg (both inclusive). If he has a total of

n

weights, each measuring an integral value in kg, then what is the minimum value that

n

can take?

Solution

Note that the weights can be placed on either side of the pan balance. For instance to measure 6 kg, we could use a 9 kg weight of the left pan and 3 kg on the right pan. This would result in measuring exactly 9 - 3 = 6 kg of items (when placed on the right pan).

Using the weights 1, 3, 9 and 27, we can measure all integers up to 1 + 3 + 9 + 27 = 40 kg.

If

3^{4}

= 81 is also used, we can measure upto 40 + 81 = 121 kg.

Therefore, minimum of 5 weights is required, i.e. 1 kg, 3 kg, 9 kg, 27 kg and 81 kg to measure any weight between 1 kg and 100 kg where the weights can be placed on either side of the pan balance.

Answer: 5

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