How many integers n, 5000 <= n <= 8000, have the property that the product of the digits is equal to 0?
+678 \(5000 \leq n \leq 8000\) (representation to LaTeX) ...
The product of the digits is always 0 if it contains at least one 0. n is a 4-digit positive integer that cannot have a 0 at the beginning.
0*** -> 0 ways
*0** -> 301 ways
**0* -> 301 ways
***0 -> 301 ways
so 0 + 301 + 301 + 301 = 903 ways. (the calculation path is from Melody)
+364 @Straight I got a different answer...
*0**
the first digit can be 5,6,7
the third digit can be 1-9
the fourth digit can be 1-9
3*9*9=243...
so 243+243+243=729?
tell me how i am wrong because i think i am
+678 uhm
500_ 10 values (0 - 10)
510_ 10 values (0 - 10)
520_ 10 values (0 - 10)
530_ 10 values (0 - 10)
540_ 10 values (0 - 10)
550_ 10 values (0 - 10)
560_ 10 values (0 - 10)
570_ 10 values (0 - 10)
580_ 10 values (0 - 10)
590_ 10 values (0 - 10)
---------------
10 * 10 = 100 values.
so when the first digit is 6 or 7, then there are (100 * 3 =) 300 values + 8000 works. So there are (300 + 1 =) 301 values.
Between 5000 and 8000 inclusive, there are [0 to 9]:
(814, 813, 813, 813, 813, 1542, 1542, 1542, 814, 813) >>>Total w/o repeats = 10319
814 integers that have at least 1 zero in them!
+364 0*00
technically it is 1-9 because if it is 0-9 then you have to -1 for 00*0 and the other ones @ straight
+364 oh and @ straight 8000 can onlly be counted once so if you are correct it is suppose to be 30300+301=901
