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For a number to be divisible by 3, the sum of each digit in the number also must be divisible by 3. Using that fact, we can narrow down the possible numbers greatly. we know that 222,333,444 are divisible by 3. and also any permutation of 2,3,4(=3x2=6) will work too since 2+3+4= 9. so a total of 9 different possibilities
I solved it in about 30 seconds. List them out!. For a number to be divisible by 3, the sum of the digits have to add up to a multiple of 3.
The first obvious number is 333.
Next, I fixed the first digit (3, as in 3XX) and brainstormed possible combinations of 2's and 4's that would add up to a multiple of 3, namely 6. The obvious combination is 2 and 4. Thus, there are at least 2 possible resulting combinations...324, and 342.
Then immediately, I noticed that any number with the combination of 3,2, and 4 are divisible by 3. Thus, more combinations include 234, 243, 423, and 432....
I stopped right there. Here are 7 possible combinations. Thus, the only possible answer is E. Whatever the other two numbers are, I just don't care...
The first obvious number is 333.
Next, I fixed the first digit (3, as in 3XX) and brainstormed possible combinations of 2's and 4's that would add up to a multiple of 3, namely 6. The obvious combination is 2 and 4. Thus, there are at least 2 possible resulting combinations...324, and 342.
Then immediately, I noticed that any number with the combination of 3,2, and 4 are divisible by 3. Thus, more combinations include 234, 243, 423, and 432....
I stopped right there. Here are 7 possible combinations. Thus, the only possible answer is E. Whatever the other two numbers are, I just don't care...
Any permutation of 2, 3, and 4...keibee82 said:
For _, _, _ ...the number of possible digits to go in the first spot is 3. The number of resulting possibilities for the second spot is 2. And 1 for the last.
Thus, the total number of possibilities is 3x2x1=6
Given that the numbers involved are exactly 2, 3, and 4 (ie. ignoring the possibilities of 222, 333, and 444). The possible permutations computed like this (or at least think of it in this way...)...keibee82 said:
There are 3 spots.... _, _, _ ....one for each number (2, 3, OR 4). And there are 3 numbers (2, 3, and 4). In the first spot, we havn't placed in any number yet. Thus, we can either put in a 2, a 3, or a 4. In other words, we can place THREE possible numbers in the first spot.
Once we place a single digit in the first spot (doesn't matter if it's a 2, a 3, or a 4), we have TWO remaining numbers. These two remaining numbers MUST go in the remaining two spots. Thus, for the second spot, we can place in either of TWO possible numbers.
Once we've placed numbers for the first and second spot, there is only ONE remaining number, and ONE remaining spot. There is only ONE way to place this remaining number.
Thus, overall ...ASSUMING WE'RE USINGS THE NUMBERS 2, 3, AND 4, the possible permutations are ...3x2x1=6.
Now, add 3 (for 222, 333, and 444).
This is the precise way to get the answer. But my way is the quick and dirty way to getting the answer. Get the answer the quickest way possible, preferably with the least amount of complex thinking! I was on my high school math team! I was top 20 in the state! ...JUST GET THE JOB DONE!!!
keibee82 said:
It's a special property of divisibility by 3. Not true for any other number. ...other than a number ending in 5 or 0 being divisible by 5.
It's a nifty fact to know.... in a number divisible by 3, the sum of its digits is also divisible by 3. ...you should add that to your arsenal of random facts to know.