We use cookies to ensure you have the best browsing experience on our website. Please read our cookie policy for more information about how we use cookies.

# Project Euler #90: Cube digit pairs

# Project Euler #90: Cube digit pairs

Contest ends in

#### Sort by

recency

#### |

#### 6 Discussions

#### |

Please Login in order to post a comment

I would make something more explicit.

One could think that it counts as 2 solutions:

This is not the case. It counts as 1 solution.

So for 9 2 the answer is 1217.

For 1 2 the answer is 13461. It can be also computed manually:

If the question had used the term

diceinstead ofcubeit would have been fun to add the input`S`

for the number of sides on the dices.A single cube only has 210 arrangements.

Hint: Generating all combinations of 3 dices works for this problem in Python. At first I thought there must be some other approach to pass all test cases, but it turns out if you find the right way to implement your combinations loop which avoids iterating through duplicate choices of dices, it would be fine.

Dice with different orders are considered as the same variant. (die1, die2) = (die2, die1).

n=10 m=3 ans=294197

Hey

When there are more than 1 cube into consideration and say for the first cube the number of possibilities are 70 and for the second cube the number of possibilities are 126.

Shouldn't the number of distinct arrangements be 70*126. I am bit stuck up at this point. Can someone give me a sample testcase for more than one cube in consideration