CAT 2001 | Question: 19

Question

In a number system the product of $44$ and $11$ is $1034.$ The number $3111$ of this system, when converted to the decimal number system, becomes

• A. $406$
• B. $1086$
• C. $213$
• D. $691$

 

Answer 1

Let the base be \(b\).

• \(44_b = 4b + 4\)
• \(11_b = b + 1\)

Their product in base \(b\) is written as $1034_b,$ i.e. \

$ (4b+4)(b+1) = 1\cdot b^3 + 0\cdot b^2 + 3b + 4.$

$\implies 4b^2 + 8b + 4 = b^3 + 3b + 4$

$\implies b^3 - 4b^2 - 5b = 0$

Factor: $b(b^2 - 4b - 5) = 0$

$ \implies b^2 - 4b - 5 = 0 \;\;\Rightarrow\;\; b = \frac{4 \pm \sqrt{36}}{2} = \frac{4 \pm 6}{2}$

So $b = 5$ or $b = -1.$ Only valid base is $b=5$ as base can only be a positive integer.

Now,

$ 3111_5 = 3\cdot 5^3 + 1\cdot 5^2 + 1\cdot 5 + 1 $

$= 375 + 25 + 5 + 1 = 406.$

Answer: 406

Answer 2

The product of $44*11$ in base-10 is $484$. now if the base is $b$ then

$(484)_{10}=(3414)_b\implies 4+b+4b^2+3b^3=484\implies 3b^3+4b^2+b=480$

now put $b=5$ it will satisfy the equation.

in decimal number system $(3111)_5\implies 1+5+25+375=406$

Option (A) is correct.

Comments

In the given question, it is given product as 1034.