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A positive integer $m$ in base $10$ when represented in base $2$ has the representation $p$ and in base $3$ has the representation $q.$ We get $p-q=990$ where the subtraction is done in base $10.$ Which of the following is necessarily true$:$

  1. $m\geq 14$
  2. $9\leq m\leq 13$
  3. $6\leq m\leq 8$
  4. $m<6$
in Numerical Ability by Boss (45.8k points)
edited by | 84 views

2 Answers

+1 vote

$ p -q = 990 \implies p = 1100, q = 110$

In base $10$ subtraction we are getting result $990$. Since $p$ is in base $2$ it can have digits $0$ or $1$ only. And since $q$ is in base $3$ it can have digits $0,1,2$ only. 

We can get result digit $0$ by

  • $0-0$
  • $1-1$

We can get result digit $9$ by

  • $10 - 1$
  • $11 - 2$

So, possible values of $p$ and $q$ can be (their decimal values must be same too)

  • $1100$ and $110$ $- 12$
  • $1101$ and $111$ $-13$

So, only matching option is Option B.

by Veteran (416k points)
0 votes

$(m)_{10}$=$(p)_{2}$= $(q)_{3}$

$\because$  p is in base 2 $\Rightarrow$ its digit can be only combination of 1 and 0.

$\because$  q is in base 3 $\Rightarrow$ its digit can be only combination of 0,1 and 2.

----------------------------------------------------------------------------------------------------------

 $(p)_{10}$- $(q)_{10}$= $(990)_{10}$

In this the last digit of p and q should be same in order to get 0.

---------------------------------------------------------------------------------------------------------------

Option B

9 $\leq$ m $\leq$ 13

$(13)_{10}$=$(1101)_{2}$= $(111)_{3}$

$\because$ last digit of 1101 and 111 is same and 1101-111 = 990

$\therefore$This could be the answer.

---------------------------------------------------------------------------------------------------------------

Option A

m$\geq$ 14

$\because$ m= 13 is satisfying

$\therefore$This could not be the answer.

---------------------------------------------------------------------------------------------------------------

Option C

6 $\leq$ m $\leq$ 8

$(8)_{10}$=$(1000)_{2}$= $(22)_{3}$

$\because$ last digit of 1000 and 22 is not same this could not be the answer.

$\because$  the remaining numbers(m=6,7) when converted to p will be less than 1000 and would be a 3 digit number which would be combination of 1 and 0 so the difference will never be $\geq$ 990 in this case.

$\therefore$ This will never be the answer.

---------------------------------------------------------------------------------------------------------------

Option D

m$<$6

$\because$  the remaining numbers(m<6) when converted to p will be less than 1000 and would be a 3 or 2 or 1 digit number which would be combination of 1 and 0 so the difference will never be $\geq$ 990 in this case.

$\therefore$ This will never be the answer.

---------------------------------------------------------------------------------------------------------------

$\therefore$ Option B.  9 $\leq$ m $\leq$ 13 is the correct answer.

by Boss (17.5k points)
0
Can there be multiple such $m$?
0
may be for m >9 but not for m<=8.

I dont know... that's why I have checked all the options for possibility.

But here we just have to eliminate options that why we don't need that information.
+1
In exam option elimination is fine. But when we give answer it is good to give the proper method unless it is so difficult.
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