Thread: Multiplication

Multiplication by 9 reverses a four-digit number (produces the four-digit number with the same digits in the reverse order.) What is the number?

I know the answer is 1089*9 = 9801 but there is a method to this madness. Does anyone know how to solve this?

Originally Posted by matgrl

Multiplication by 9 reverses a four-digit number (produces the four-digit number with the same digits in the reverse order.) What is the number?

I know the answer is 1089*9 = 9801 but there is a method to this madness. Does anyone know how to solve this?

Solve what? You just did it. Do you mean to prove there's no other possible answer?

yes please and thank you i do not know how to prove this

Why not try something like setting up the following equations:

9(a + 10b + 100c + 1000d) = d + 10c + 100b + 1000a (digit reversing)
a+b+c+d = 9n, for some integer n (divisibility by 9). Either n = 1, 2, 3, or 4, because of the following constraints:

0 <= a,b,c,d <= 9.

Maybe that could get you started?

Hello, matgrl!

I will assume that are four different digits.


. .


In column-1, we see that: .
And there is no "carry" from column-2.


So we have:

. .


Since there is no "carry" frim column-2, must be 0 or 1.
Since , then: .


Then we have:

. .


In column-3, we have: . ends in
. . Hence, ends in


Therefore, the solution is:

. .

wonderful thank you!