Problem 6

The 4-digit numbers x and y have the property that the digits of y are the digits of x in reverse order. Assume the first digits are not zero. If z = x + y, how many different values of z are greater than 10,000?

[Problem submitted by S. Lee, LACC Professor of Mathematics.]

Solution for Problem 6:

If x has the 4 digits abcd, then y has the 4 digits dcba.

Then z=(1000a+100b+10c+d)+(1000d+100c+10b+a)=1001(a+d)+110(b+c).

If z>10000, and a+d=9, then b+c must be 10,11,12 … or 18. That gives 9 different values for z.

If z>10000, and 10 a+d18, then 0b+c18. That gives 9 values for a+d and 19 values for b+c, therefore 9x19 values for z.

Together, there are 9+9x19=180 different values for z.