Let N be a 3-digit positive integer and let M be another 3-digit positive integer obtained by reversing the digits of N.

Find the number of possible values of N such that N−M=396.

Guest Jul 30, 2020

#1**+1 **

Here is what I** THINK**:

xyz - zyx = 396

100x + 10 y + z - 100z - 10 y - x = 396

99x - 99z = 396

99(x-z) = 396

x-z = 4

x can be 9 8 7 6 5

z would be 5 4 3 2 1

y can be anything 0-9

5 x 10 = 50 edited I had 4 as a possible for x but it isn't a possible

ElectricPavlov Jul 30, 2020

#2**0 **

EP: That is correct, provided they accept revesal of 10 of those numbers that end in zero, such as:

400 - 004 =396, 450 - 054 =396....etc. if they are not acceptable, then there are only 50 such numbers.

Guest Jul 30, 2020

#3**0 **

Hmmmm.... I do not have any of the numbers ending or starting with zero..... Only the middle number can be zero in my answer....

I still find 60 numbers..... can you explain further?

Wait a minute.....I found my error ....I edited my answer above.....I mistakenly listed 4 as a possible, when I should have only counted down to 5 ...just a math error in my post.... THANX !

9 n 5 - 5 n 9

8 n 4 - 4 n 8

7 n 3 - 3 n 7

6 n 2 - 2 n 6

5 n 1 - 1 n 5 'n, can be any 0-9 so 5 x 10 = 50

ElectricPavlov
Jul 30, 2020