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The question is from Permutation and Combination. This question combines number theory with combinatorics. We need to find out the number of multiples of 12 which comprises of only 2's and 3's. This section hosts a number of questions which are on par with CAT questions in difficulty on CAT Permutation and Combination, and CAT Probability.

Question 5: A seven-digit number comprises of only 2's and 3's. How many of these are multiples of 12?

  1. 11
  2. 12
  3. 10
  4. 22

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Explanatory Answer

Method of solving this CAT Question from Permutation and Combination: One unit number theory and one unit Combinatorics makes a potent combination for fabulous questions!

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.
All seven 3's - No possibility.

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 = \\frac{5!}{3!2!}\\) = 10

Six 2's and one 3 - The first 5 digits should all be 2's. So, there is only one number 2222232.
So, there are a total of 10 + 1 = 11 solutions.

The question is "A seven-digit number comprises of only 2's and 3's. How many of these are multiples of 12?"

Hence the answer is "11"

Choice A is the correct answer.

 


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