Question Number 178338 by Acem last updated on 15/Oct/22

Commented by Acem last updated on 17/Oct/22

Answered by mr W last updated on 16/Oct/22
![[A] ⌊((9999)/8)⌋−⌊((999)/8)⌋=1249−124=1125 solution for A is simple. we just need to count the number of terms of an A.P.: 1000, 1008, 1016, ....,9992 that′s ((9992−1000)/8)+1=1125 numbers.](https://www.tinkutara.com/question/Q178365.png)
Commented by mr W last updated on 16/Oct/22

Commented by mr W last updated on 16/Oct/22
![[B] say such a number is ABCD with A≠B≠C≠D, A≠0 determinant (((CD=),(08),(16),(24),(32),(40),(48),(56),(64),(72),(80),(96)),(B_n ,3,(3/1),(2/1),(3/1),3,(2/1),4,(2/1),(3/1),3,(3/1)),(A_n ,7,(6/7),(6/7),(6/7),7,(6/7),(6/7),(6/7),(6/7),7,(6/7)),(,(21),(25),(19),(25),(21),(19),(25),(19),(25),(21),(25))) 21×3+25×5+19×3=245 determinant (((CD=),(04),(12),(20),(28),(36),(52),(60),(68),(76),(84),(92)),(B_n ,5,4,5,5,4,4,5,5,4,5,4),(A_n ,7,6,7,6,6,6,7,6,6,6,6),(,(35),(24),(35),(30),(24),(24),(35),(30),(24),(30),(24))) 35×3+24×5+30×3=315 ⇒totally 245+315=560 numbers with different digits ================ explanation of the table say the number is ABCD. when CD=08, B can be 2, 4, 6, i.e. 3 possibilities (B_n =3). for each B=2, A can be 1,3,4,5,6,7,9, i.e. 7 possibilities (A_n =7). ⇒3×7=21 numbers.](https://www.tinkutara.com/question/Q178368.png)
Commented by Acem last updated on 16/Oct/22

Commented by Acem last updated on 16/Oct/22

Commented by mr W last updated on 16/Oct/22

Commented by Acem last updated on 17/Oct/22

Commented by mr W last updated on 18/Oct/22

Commented by Acem last updated on 18/Oct/22

Answered by Acem last updated on 19/Oct/22

Commented by Acem last updated on 16/Oct/22

Answered by Acem last updated on 19/Oct/22
