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?
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:
. .