Register

ISI2019-MMA-4

• edited by
2,039 views
1 votes
1 votes

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. $360$
• edited by
User Avatar

1 Answer

Best answer
4 votes
4 votes
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
• selected by
User Avatar
Voting & Accepting an answer helps improve the community
← PreviousNext →
← Previous in categoryNext in category →

Related questions

0 votes
0 votes
3 answers
2
Sayan Bose asked May 7, 2019
2,513 views
ISI2019-MMA-20
Suppose that the number plate of a vehicle contains two vowels followed by four digits. However, to avoid confusion, the letter $‘O’$ and the digit $‘0’$ are not used in the same number plate. How many such number plates can be formed? $164025$ $190951$ $194976$ $219049$
Suppose that the number plate of a vehicle contains two vowels followed by four digits. However, to avoid confusion, the letter $‘O’$ and the digit $‘0’$ are not ...
User Avatar
0 votes
0 votes
1 answer
3
Sayan Bose asked May 5, 2019
3,436 views
ISI2019-MMA-2
The number of $6$ digit positive integers whose sum of the digits is at least $52$ is $21$ $22$ $27$ $28$
The number of $6$ digit positive integers whose sum of the digits is at least $52$ is$21$$22$$27$$28$
User Avatar
4 votes
4 votes
1 answer
4
Sayan Bose asked May 7, 2019
1,805 views
ISI2019-MMA-30
Consider the function $h$ defined on $\{0,1,…….10\}$ with $h(0)=0, \: h(10)=10 $ and $2[h(i)-h(i-1)] = h(i+1) – h(i) \: \text{ for } i = 1,2, \dots ,9.$ Then the value of $h(1)$ is $\frac{1}{2^9-1}\\$ $\frac{10}{2^9+1}\\$ $\frac{10}{2^{10}-1}\\$ $\frac{1}{2^{10}+1}$
Consider the function $h$ defined on $\{0,1,…….10\}$ with $h(0)=0, \: h(10)=10 $ and$$2[h(i)-h(i-1)] = h(i+1) – h(i) \: \text{ for } i = 1,2, \dots ,9.$$Then t...
User Avatar
Link Copied!