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Consider the equation $(123)_5=(x8)_y$ with $x$ and $y$ as unknown. The number of possible solutions is _____ .
asked | 1.9k views
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why (5,6) and (6,5) are not the possible pair n why ? the minimum base should be greater than 8 ? can someone expain ?
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the minimum base should be greater than 8 ? can someone expain ? n why (5,6) (6,5) are not possible pairs
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@priyanka gautam 1

here (x8)y means y>8 means from 9 onwards..
also x<y
therefore,(5,6) is not possible because y is not >8
and (6,5) is not possible because x is not <y...

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ok got it x<y but why y> 8 ? plz expalin ?
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since (x8)..one number here is 8 so base has to be more than 8
for base 2 numbers are 0,1
base 3..0,1 2
and similarly if a number is 8...its base has to be more than 8..
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thnku

Converting both sides to decimal,

$25+10+3=x$*$y+8$

So, $xy=30$

Possible pairs are $(1,30),(2,15),(3,10)$ as the minimum base should be greater than $8.$
answered by Junior (697 points)
edited
+9

And also x<y (strictly). Otherwise pair (30,1), (15,2) can also be included.

Changing (123) base 5 into base 10= 1*25+2*5+3*1=38 Changing x8 base y in decimal= x*y+8 Equating both we get xy+8=38

• xy=30
• possible combinations =(1,30),(2,15),(3,10)

but we have ‘8’ present in x8 so base y>8 as all three are satisfying the conditions so total solutions =3 hence ans is ( C) part

answered by Loyal (8.8k points)
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why not 5,6 plz someone explain properly ..
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8 can't be part of digits if we take  5,6
here the equation is x*y=30 so the possible values for x and y is (5,6) other values like (6,5) (3,10)(10,3) etc are not valid.
answered by Active (1.2k points)
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How can 5,6 be valid when the number with base x is 8?
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vlaue of y never less then 8 show (5,6) is also invalid

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