Permutation-Combination
Aptitude

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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.

$47$

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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:

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

$= 0+10+1$

$= \textbf{11}$


(2) Comment(s)


Prashanth
 ()

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


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?