Friday

April 18, 2014

April 18, 2014

Posted by **urgent answer needed** on Saturday, June 15, 2013 at 4:37am.

Details and assumptions

The last stretch that the dog runs must also satisfy the condition on the minimum number of consecutive blocks.

An intersection is reachable if the dog runs through it. It doesn't matter if the dog can change direction at that intersection.

- math -
**MathMate**, Saturday, June 15, 2013 at 7:22amAll of them.

Since there are 46x46 streets, there are 45x45 blocks.

The dog has to run 4 blocks in a row in the horizontal direction, and 12 blocks in a row in the vertical direction, both of which are even numbers, it will never reach the food, in which direction it tries.

- math -
**rohit**, Saturday, June 15, 2013 at 10:29amthen wats the answer

- math -
**MathMate**, Saturday, June 15, 2013 at 2:45pmI guess I did not interpret the question correctly the first time around.

"At least 4 blocks" means that the 2nd, 3rd, 4th streets (north-south) are not accessible.

Similarly, the 2nd to the 12th avenues (east-west) are not accessible.

So near the starting point there are 46×11 intersections not accessible, in addition to the 2nd, 3rd and 4th streets of the first avenue,

for a total of

46×11+3=509 intersections.

Similarly, for the end point (top-right), the same number applies.

The total number of inaccessible intersections is therefore 2×509=1018.

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