Bunuel wrote:
How many seven-digit positive integers are there,which have only 2's and/or 3's as the digits and which are multiples of 12?
A. 10
B. 11
C. 12
D. 13
E. 14
Solution:In order to be a multiple of 12, it must be a multiple of 3 and also a multiple of 4. Since the divisibility rule for 4 is that the last two digits of the number form a number that is divisible by 4, we see that the last two digits of such a 7-digit number must be 32 (since 23, 22 and 33 are not divisible by 4). Furthermore, since the divisibility rule for 3 is that the sum of the digits of the number is divisible by 3, we see that the sum of the first five digits and 5 (which is the sum of the last two digits) must be a multiple of 3. Since the least sum of the first five digits is 2 x 5 = 10 and the greatest sum of the first five digits is 3 x 5 = 15, the sum of the first five digits must be 10 or 13 so that when added to 5, it’s a multiple of 3.
If the sum of the first five digits is 10, then all the digits must be 2, i.e., 22222. We see that there is only one such 7-digit number in this case, i.e., 2222232.
If the sum of the first five digits is 13, then three digits are 3 and two digits are 2, e.g., 33322. However, there are 5!/(3! x 2!) = 10 ways to arrange three 3s and two 2s. Therefore, there are 10 such 7-digit numbers in this case.
Therefore, altogether there 1 + 10 = 11 such 7-digit numbers.
Answer: B
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