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 $4$ to $6$. 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 $a_1$ to $a_{n}$.

In fact, $a_{1} = 2^{0}$ to $a_{n} = 2^{n-1}$ 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 $3^{0}$ to $3^{n - 1}$, 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