Entries in An Array
⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎡ 2 0 1 0 1 0 0 2 0 1 1 0 1 0 2 0 1 0 0 1 0 2 1 0 2 2 2 2 2 0 0 0 0 0 0 0 ⎦ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎤
In the above 6 × 6 array, one can choose any k × k subarray, with 1 < k ≤ 6 and add 1 to all its entries. Is it possible to perform the operation a finite number of times such that all entries in the array are multiples of 3?
Note: If yes, state the number of times in which the operation is required. If no, type 0 .
Post your solutions so others may learn! Thanks!
The answer is 0.
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3 solutions
Maybe I don't understand the question. It doesn't say anything about multiplying or scaling, but it says adding 1. Also it says you can choose any square sub array, not just limited to 2x2. Therefore wouldn't one solution be to choose k=1 and perform the operation on each entry as many times as needed to make every entry equal to 3? Is there some other constraint or requirement that I'm missing? Appreciate some clarification.
Any multiple of three multiplied by another multiple of three would be greater than six
In the k × k array, we can see we can't make every number a multiple of 3 unless the top right '0' becomes a multiple of 3. We can only make the top right '0' a multiple of 3 when k = 6 in which every number will change. Plugging k = 6 into the matrix and it becomes like such: ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎡ 3 1 2 1 2 1 1 3 1 2 2 1 2 1 3 1 2 1 1 2 1 3 2 1 3 3 3 3 3 1 1 1 1 1 1 1 ⎦ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎤
Now the top right '0' has become 1. The fastest way to get that top right '0' to a multiple of 3 is to use k = 6 3 times. But now the whole top row changes, from (3 1 2 1 3 1) to (5 3 4 3 5 3) . We can now see no matter how many times we add 1 to the original top right '0' and top left '2' , they will never simultaneously be a multiple of 3. Hence, it is impossible and the answer is 0 .
There is no 2x2 subarray that does not contain 0, which obviously requires 3 to be a multiple of 3. Scaling 1 or 2 with 3 will never result in a multiple of 3, so the answer is 0.