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# number theory

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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.

Jul 30, 2020

#1
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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

Jul 30, 2020
edited by ElectricPavlov  Jul 30, 2020
#2
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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.

Jul 30, 2020
#3
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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
edited by ElectricPavlov  Jul 30, 2020