The Gateway to Computer Science Excellence
First time here? Checkout the FAQ!
x
0 votes
204 views

Suppose that $6$-digit numbers are formed using each of the digits $1, 2, 3, 7, 8, 9$ exactly once. The number of such $6$-digit numbers that are divisible by $6$ but not divisible by $9$ is equal to

  1. $120$
  2. $180$
  3. $240$
  4. $360$
in Combinatory by Loyal (7k points)
edited by | 204 views

1 Answer

+2 votes
Best answer
test of divisibility by 9 is: sum of digits is divisible by 9.

here sum of digits is 1 + 2+ 3+ 7+ 8+9 = 30

so all numbers formed by these digits exactly once are not divisible by 9.

But all numbers formed by these digits are divisible by 3.

so we have to see the number of numbers which are even.

there are totally 6! = 720 numbers possible with these digits.

Out of which 1/3rd are even(2,8 are even digits out of 6 available digits)

So the answer is 240.

C
by Active (2k points)
selected by

Related questions

Quick search syntax
tags tag:apple
author user:martin
title title:apple
content content:apple
exclude -tag:apple
force match +apple
views views:100
score score:10
answers answers:2
is accepted isaccepted:true
is closed isclosed:true
49,845 questions
54,784 answers
189,430 comments
80,448 users