+1 vote
148 views
consider the number:

5324+2342=(7666) of base r

what is the value of r:

a.8

b.10

c.16

d.>=8
asked
edited | 148 views
0
I am not getting whether lhs is in base r or it is in decimal(base=10)
0
actually LHS the number is given in any radix(8,10,16 or greater then 8) and the ans is given is in the same radix which is used in LHS. we have to find out which one is correct???
0
I think, the value of r has to be less than 8 (and that means, no option is correct).

Here, min radix is 8. Using radix 8, we get (5324)+(2324) = (7650), but the RHS has (7666), so, the radix r has to be less than 8 - and that will make digit 7 invalid !
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by using hit and trial no any option is correct.
0
Actually, I correct the question. Now check it.
0
Radix 8 works now.

## 3 Answers

+1 vote
Let's check one by one option,

${\color{Blue} {(a)}}$   r=8,when we add LSB bits of both (4+4=8) should be zero and one carry but at RHS, LSB bit is 6 .so option a is wrong.

${\color{blue}{(b)}}$   r=10,it is simply when we add LSB bits of both then answer should be 8 but at RHS, LSB bit is 6 so option B is also wrong.

${\color{blue}{(c)}}$. r=16,in this also when we will add LSB bits then LSB bit will be 8,same reason as above option ,,so this is also wrong.

${\color{blue}{(D)}}$,  when at r=8,10,16 is not satisfied then r>=8 is also false (only one condition is enough to make false)

So none of these option is true.
answered by Boss (10.1k points)
0 votes
r must be greater than 7

(5324)base r +(2324) base r=(7666) base r

(5*r^3+3*r^2+2*r+4)+(2*r^3+3*r^2+2*r+4)=(7*r^3+6*r^2+6r+6)

2r=2

r=1(rejected)

hence no value of r for which it is satisfied
answered by Active (3.9k points)
0 votes
I am not getting whats wrong with u people

Its direct base 10  ..
answered by (165 points)

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