Pick a number
4 6 1 0 1 6 1 8 2 0 6 9 1 5 2 4 2 7 3 0 1 0 1 5 2 5 4 0 4 5 5 0 1 4 2 1 3 5 5 6 6 3 7 0 2 2 3 3 5 5 8 8 9 9 1 1 0 2 6 3 9 6 5 1 0 4 1 1 7 1 3 0
Choose 6 numbers from the given table, in such a way that from each row and column, exactly 1 number is chosen.
What is the probability that the product of these numbers is divisible by 6 5 ?
If the answer is p , give your answer as ⌊ 1 0 0 0 p ⌋ .
The answer is 0.
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
1 solution
An easier way is to get the product of a diagonal in the matrix because that is the only way for a number in the matrix to be picked in a different row and column each time. Luckily, both diagonals product are the same. Thus, the product of both of the diagonals is 648648000 and when you divide it by 6^5 it is 83416.6667, which isn't an integer, and thus, the probability that the product of these numbers is divisible by 6^5 is 0%
Log in to reply
Check your thinking again as there are many other ways than the diagonals themselves.
If you see the matrix carefully, each row can be viewed as a product of number 'n' and numbers 2 ,3,5,7,11,13 respectively . The number ' n' for row First is 2 For second row it is 3 For the third row it is 5 Fourth one ,it's 8 Fifth and sixth one is 9 and 10 respectively. Now if you need to select any number abiding by the conditions given The product would be 2×3×5×8×9×10×2×3×5×7×11×13. And now if one can see the highest power of 3 in the above product is 4. Our devisor is having highest power of 5. So the product would never be divisible. Hence probability would be zero irrespective of the order of selecting the numbers.