The number of numbers greater than 5000 , less than 9000 and divisible by 3 , that can be formed using the digits 0,1,2,5,9, if the repetition of the digits is allowed, is ____
Answer: 42
Solutions · 1
Official solution
To find numbers of number greater than 5000 and less than 9000 so number will start with 5 or 9 of the form 5 abc and a,b,c∈0,1,2,5,9 also the number is divisible by 3 .
This means 5+a+b+c is divisible by 3
so the sum a+b+c is of the type 3n+1.
unit place c determined by the sum of previous 3 places.
rem(a +b)= remainder of a+b
rem(a+b)
combination (a,b)
value of c
total = combination (a, b) \times c(choice)
(0, 0), (0, 9), (9, 0), (9, 9), (1, 2), (2, 1), (5, 1), (1, 5)
8 \times 1 = 8
(0, 1), (9, 1), (1, 0), (1, 9), (2, 2), (2, 5), (5, 2), (5, 5)
0, 9
8 \times 2 = 16
(0, 2), (0, 5), (9, 2), (9, 5), (2, 0), (2, 9), (5, 0), (5, 9), (1, 1)
2, 5
9 \times 2 = 18
Total = 8+16+18=42