Question 1193916
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The first digit can't be 0; and there are special requirements if the first digit is 3.  So consider two cases -- first digit 1, 2, 4, or 5; or first digit 3.<br>
(1) first digit 1, 2, 4, or 5<br>
There are 4 choices for that first digit, then there are no restrictions for the other digits; so the number of 4-digit numbers with first digit 1, 2, 4, or 5 is 4*6*6*6=864.<br>
(2) first digit 3<br>
The first digit has to be 3 (1 choice); the last digit must be odd (1, 3, or 5 -- 3 choices); and there are no restrictions on the other two digits.  The number of possible 4-digit even numbers with first digit 3 is 1*3*6*6=108.<br>
The total number of 4-digit numbers with the given requirements is 864+108=972.<br>
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A comment for tutor @ikleyn, whose math is nearly always very good but whose English is often not....<br>
There is nothing deficient in the statement of the problem.  It says that "each digit of the number is either 0,1,2,3,4 or 5".  That means repetition of digits is allowed.<br>