digits of a 3 digit no. in base 11get reversed,when expressed in base 9 find all such no.s in base 11.?

Question

options given are: 243,467-first;302,604-second;203,406-third;203,406,609-fourth

Answer 1

let N="xyz" be the 3-digit number...in base 11

so N = z + 11y + 121x (decimal sum)

also N = x + 9y + 81z (reversed in base 9)

equating both we get

120x + 2y = 80z or 60x+y=40z

or y = 20(2z-3x) which means that y=0 since 2z-3x>=0

hence we reduce the equation to 60x=40z or 3x=2z which has the solutions as x=2,z=3 or x=4, z=6 or x=6, z=9

hence 203, 406 and 609 are the numbers in base-11; reversing expresses them in base-9

note: base 11 includes an additional symbol, apart from 0 to 9; we have ignored that in this problem

Answer 2

Let the number be "xyz" in base 11 so 121x+11y+z
when digits are reversed yxz in base 9 = 81z+9y+x
x,y,z < 9(as base 9 is used

121x+11y+z = 81z+9y+x
or 120x-2y-80z = 0
y has to be zero else minimum 20 to be dvisible by 40
so 120x = 80 z or 3x = 2z
so values are

x =0 z = 0 number 000
x=2 z = 3 203(base 10 value 245)
x= 4 z = 6 406(base 10 490
these are only 3 solutions

Answer 3

Let the number be [xyz] in base 11, then it is [zyx] in base 9. Here, 0 <= x, y, z <= 8.

121x + 11y + z = 81z + 9y + x
120x + 2y - 80z = 0

Since 120x and 80z are multiples of 40, 2y must also be a multiple of 40; therefore, y = 0. We are left with

3x - 2z = 0

with possible solutions

A) x = 0; z = 0
B) x = 2; z = 3
C) x = 4; z = 6

The three numbers are

A) 000[11] = 000[10] = 000[9]
B) 203[11] = 245[10] = 302[9]
C) 406[11] = 490[10] = 604[9]