Aptitude Discussion

Q. |
A seven-digit number comprises of only 2's and 3's. How many of these are multiples of 12? |

✖ A. |
$1$ |

✔ B. |
$11$ |

✖ C. |
$21$ |

✖ D. |
$47$ |

**Solution:**

Option(**B**) is correct

Number should be a multiple of 3 and 4.

So, the sum of the digits should be a multiple of 3.

We can either have all seven digits as 3, or have three 2's and four 3's, or six 2's and a 3 (The number of 2's should be a multiple of 3).

For the number to be a multiple of 4, the last 2 digits should be 32. Now, let us combine these two.

(i)- All seven 3's - No possibility

(ii)- Three 2's and four 3's - The first 5 digits should have two 2's and three 3's in some order. No of possibilities $= \dfrac{5!}{3!×2!} = 10$

(iii)- Six 2's and one 3 - The first 5 digits should all be 2's.

So, there is - only one number - 2222232

So we get the total of seven-digit number comprises of only 2's and 3's which are multiples of 12 are:

$= (i) + (ii) + (iii)$

$= 0+10+1$

$= \textbf{11}$

**Prashanth**

*()
*

in case 2: Here in ending one 3,2 is fixed for dividing by 4 so, remaining digits are two 2's and three 3's

result is 6!/2!3!

is this wrong?

but in case 2 where it checked which is divisible by 4 or not?