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Odd and Even 3 (Posted on 2016-04-04) Difficulty: 3 of 5
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.

No Solution Yet Submitted by K Sengupta    
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Solution My answers (with and without leading zeros) | Comment 3 of 6 |

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

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