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# In the decimal system of numeration the number of 6-digit numbers in which the sum of the digits is divisible by 5 is (A) 210(B) 84(C) 126(D) none

Vikas TU
14149 Points
4 years ago
So any of the remaining 9 digits can fill the first place. So no. of ways to fill first digit =9,
The next four places can be filled with any of the digits. So no. of ways to fill next 4 digits = 10 each. For four digits it is 104.
Varun Patkar
12 Points
3 years ago
The options given by you are wrong. The options are a)180000 b)540000 c)500000 and d)none.The solution isLet the sum ot`any t`ive digits be N. Now N can be any number from 1 to 45. In each case the remaining digit can be chosen in two particular ways only so as sum of digits is divisible by 5. {e.g. ir N-42. Then remaining digit can be 3 or 8: if N=23. Then remaining digit is 2 or 7;... Etc Hence total possible number of such numbers is 9 × 104 × 2 I.e. 180000.
Varun Patkar
12 Points
3 years ago
Now there are 90,000 5 digit #s Each such # can be made divisible by 5 by adding the sixth digit in 2 ways, eg • if the 5 digit # leaves remainder 0 when divided by 5, the sixth digit has to be 0 or 5 • if the remainder ........ is 1, the sixth digit has to be 4 or 9 and so on & so forth Thus answer = 90,000 *2 = 180,000