CAT 2025 Lesson : Number Systems - Arithmetic Operations

In the base $n$ number system, the product of $223$ and $43$ is $12031$. What is the value of $n$?
(1) $6$           (2) $7$           (3) $8$           (4) $9$          

Solution

The product of units digits in base $10$ is $9$. However, the units digit in base $n$ is $1$.

$9$ cannot be written as $1$. So, $\bm{9}$ probably equals $\bm{11, 21, 31,}$ etc. in the base $\bm{n}$ system.

If $9$ in decimal system is $11$ in base $n$ system, then
$1 \times n + 1 = 9$
⇒ $n = 8$

If $9$ in decimal system is $21$ in base $n$ system, then
$2 \times n + 1 = 9$
⇒ $n = 4$ (Not possible as $4$ is used as a digit in $43$ and answer options do not have $4$)
(For numbers $31, 41,$ etc., $n$ becomes smaller than $4$, which is not possible.)

∴ $n = \bm{8}$

Answer: (3) $8$

In a certain number system, the sum of $4042$ and $1421$ is $11013$. The sum of these numbers when written in the decimal system is

Solution

The units digits are equal in both bases. However, the tens digits are different.

The sum of these numbers, if they were in base $10$, is $(5463)_{10}$. This is lower than $(11013)_{n}$. Therefore, $n$ has to be less than $10$.

So, $(4 + 2 = 6)$ in base $n$ probably yields $11, 21,$ etc., wherein $1$ is written down and $1, 2,$ etc. is carried over respectively for the next digit.

If $6$ in decimal system is $11$ in base $n$ system, then
$1 \times n + 1 = 6$
⇒ $n = 5$

If $6$ in decimal system is $21$ in base $n$ system, then
$2 \times n + 1 = 6$
⇒ $n = 2.5$(Not possible)
(Other smaller values for $n$ are not possible.)

∴ $n = 5$

$(11013)_{5} = (1 \times 5^{4}) + (1 \times 5^{3}) + (1 \times 5) + 3 = 758$

Answer: $758$

What is the sum of $(235)_{7}$ and $(326)_{7}$ in the base $7$ system?

Solution

$(235)_{7} = (2 \times 49) + (3 \times 7) + 5 = 124$

$(326)_{7} = (3 \times 49) + (2 \times 7) + 6 = 167$

$124 + 167 = 291$



$291 = (564)_{7}$

Answer: $564$