Question

Find the number of possible three digit number in base 11 when expressed in base 9 has its digits reversed.​

Answer 1

Answer:

Take any three-digit positive integer, reverse its digits, and subtract. For example, 742 − 247 = 495. The difference is divisible by 11. Prove that this must happen for all three-digit numbers abc

Step-by-step explanation:

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Answer 2

Answer:

The number is verified

Step-by-step explanation:

Let's suppose three digits in the base 11 system are x, y & z

Base 11 System:

One's place - x

11s place - y

121s place - z

So, the number formed $= 121z + 11y + x------(1)$

Now, the digits in Base 9 System z, y, x

Base 9 system :

One's place - z

9s place - y

81s place - x

So, the number formed $= 81x + 9y + z-----(2)$

As per the condition, these two numbers are equal

So,$121z + 11y +x = 81x + 9y + z$

$= > 121z-z+11y-y+x-81x= 0$

$= > 120z+10y-80x=0$

divide the above equation by 10

$= > 12z+y-8x=0$

$= > y=8x-12z=0$

$== > y=8(x-4z)=0$

$= > y = 8(x - 4z)=0$

Now, y is the common digit of the number formed in both the systems. And it comes in the middle place. Since y is a digit, it can hold value either

0, 1,2,3,4,5,6,7,8,or 9,10.

Since, y = 8 times (x--4z)

ie, 8*0 =0

8*1 = 8

So these 0 & 8 are the possible values of digit y. But in Base 9 system we have digits 0 through 8 . So y = 8 is ruled out

That concludes that y = 0

=> x-4z=0

=> x=1 z=4 => z =4 & x=1

We get , x = 1, y= 0, z = 4

So the number in base10 system is 104 & the sum of its digits = 5 . . . . . . . . . . Ans

VERIFICATION:

So, the number in Base 11 System= 121*4+ 11*0 + 1=485

&the number in Base9 System = = 81*1 + 9*0 + 4=85

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