# A positive integer N is in base 5, a 3 digit number. When 3N is in base 9, it's 3 digit number consists of same digits same order as 1st. Find decimal representations of all such numbers N.

### 3 Answers | Add Yours

So lets say N = `xyz_5`

Then N = `x*5^2+y*5^1+z^5^0` = 25x+5y+z

So in decimal we can write;

N = 25x+5y+z-----(1)

3N = `xyz_9`

3N = `x*9^2+y*9^1+z*9^0`

3N = 81x+9y+z---(2)

(1)*3 = (2)

75x+15y+3z = 81x+9y+z

6x-6y-2z = 0

z = 3x-3y

Since N is in base 5 x,y or z cannot have a value beyond 4. Also x should be greater than 0 since it is 3 digit number and it x,y,z should satisfy the equation z = 3x-3y

`0ltxlt=4`

`0lt=ylt=4`

`0lt=zlt=4`

z = 3x-3y

let z = 0 then x=1 and y=1

let z = 1 then x= 1 and y=0

let z = 2 then x or y have no answers

let z = 3 then x = 2 and y =1 OR x=3 and y = 2 OR x=4 and y=3

let z = 4 then x and y have no answers.

So the numbers are;

`110_5`

`101_5`

`213_5`

`323_5`

`433_5`

In decimal;

**30**

**26**

**58**

**88**

**118**

In base 9;

`110_9` = 90 = 3*30

`101_9` = 82 = 3*26

`213_9` = 174 = 3*58

`323_9` = 264 = 3*88

`433_9` = 354 = 3*118

26 wont work- 26*3=78. 78 is not higher than 81, wont work.

for jeew-m's answer, it all makes sence bar the rest. when you factor in the values for z, it doesn't work!

ie: let z=1 then x=1 and y=0

this uses the equation z=3x-3y

how is 1=3(1)-3(0)

ie. 1=3-0

True! its not! yet how did it become a correct answer at the end...?

doesnt make sence to me.. if anyone can explain it! thanks