Easy Question

[da_silvy]

More or less a quote from a puzzle...
Suppose the hundred's digit of a three digit number is greater than the 10's digit and greater than the units digit.

When the digits in this number are reversed, and the result is subtracted from the original number, the units digit of the difference is 4.

a) What is the difference between the 3 digit number and its reverse.

b) How many different 3 digit numbers meet the give conditions?




Please help me...

[unformed]

a) 594
b) 15
----------------------------


the way i came up with it:

you have a number (each letter is a digit)
ABC
where A>B>C

you know that:
ABC
-CBA
----
DE4

since A>C, here are the possiblities for A & C:
A=9 C=3
A=8 C=2
A=7 C=1

[assume => means must equal]
if A=9, C=3, then B => {4...8}
if A=8, C=2, then B => {3...7}
if A=7, C=1, then B => {2...6}

trying the numbers out
943-349=594
983-389=594
832-238=594
...

then just count the different numbers, there's fifteen total

[da_silvy]

There's 30 combinations, i worked out

If A=9, C=3
983, 973, 963, 953, 943, 933, 923, 913, 903

(That's 9 combos)

If A=8, C=2
872, 862, 852, 842, 832, 822, 812, 802

(That's 8 combos)

If A=7, C=1
761, 751, 741, 731, 721, 711, 701

(That's 7 combos)

If A=6, C=0
650, 640, 630, 620, 610, 600

(That's 6 combos)

9 + 8 + 7 + 6 = 30
17 + 13 = 30