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How many integers n, 5000 <= n <= 8000, have the property that the product of the digits is equal to 0?

 Jan 7, 2022
 #1
avatar+675 
+1

\(5000 \leq n \leq 8000\) (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)

 Jan 7, 2022
edited by Straight  Jan 7, 2022
 #2
avatar+360 
+2

@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

 Jan 7, 2022
edited by XxmathguyxX  Jan 7, 2022
 #4
avatar+675 
+1

*0**

 

The first digit can be 5, 6, 7           3 values

The second digit must be 0           1 value

The third digit can be 0-9             10 values

The fourth digit can be 0-9           10 values

 

3 * 1 * 10 * 10 = 300.

 

8000 is also a possible value.       +1

 

300 + 1 = 301 values.

Straight  Jan 7, 2022
 #3
avatar+675 
+1

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.

 Jan 7, 2022
 #5
avatar+675 
0

wait wait wait, why do have a different answer...?

 Jan 7, 2022
 #6
avatar
+1

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!

 Jan 8, 2022
 #7
avatar+360 
+1

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

 Jan 9, 2022
 #8
avatar+360 
+1

melody can you help us?

 Jan 9, 2022
 #9
avatar+360 
+1

oh and @ straight 8000 can onlly be counted once so if you are correct it is suppose to be 30300+301=901

 Jan 9, 2022

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