Determine the total number of values of a positive integer N having at most four digits such that:

(i) The digit 1 occurs an odd number of times in N, and:

(ii) Each of the digits 2, 0 and 6 occurs an even number of times in N. (An even number includes zero), and:

(iii ) Any of the remaining six digits may or may not occur in N.

Since the issue of the leading zeros was not addressed  (despite my remarks on the reviewing board)  we will provide 2 solutions   for the price of one,    first  allowing numbers like 0081 or 0170 then discounting the appropriate quantity from the inclusive total.

Let the letter a stand for any member of (0,2,6),  and b for any member of (3,4,5,7,8,9) .

Now let’s    count all number below 10000 that comply with puzzle’s specs:

1digit      number:   1                                                                                       qty     1

2dig        numbers: 1b,b1,  6-each
                                              qty     12        
  

3dig        numbers:   111(1),   aa1(3*3), 
  bb1 (3*6*6)                                  qty     118

4dig        numbers   111b(4*6),     
    aa1b(4*3*6*3) ,  bbb1 (4*6*6*6)                        qty     24+5*216=1104

 

Total  131+1104=1235      ;              if I did not err..

Discounting  010 (2),   00b1(36),
 i.e.   total 38       leaves us with  1197      complying numbers.

 

ANSWERS:   1235 (WITH LZ  )
&   1197 (WITH NO LZ  )     

Edited on April 5, 2016, 4:24 pm

Posted by Ady TZIDON on 2016-04-05 07:15:40