BdMO National 2021 Higher Secondary Problem 5

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Anindya Biswas
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BdMO National 2021 Higher Secondary Problem 5

Unread post by Anindya Biswas » Sun Apr 11, 2021 9:33 pm

তুমি কতভাবে তিনটা \(20\)-তলবিশিষ্ট ছক্কা মারতে পারবে যাতে ছক্কাগুলোতে ওঠা সংখ্যাগুলোর যোগফল ঠিক \(42\) হয়? এখানে কী কী সংখ্যা উঠছে, তার ক্রম গুরুত্বপূর্ণ।
(মনে রেখো যে একটা \(20\)-তলবিশিষ্ট ছক্কা একটা সাধারণ ছয়তলবিশিষ্ট ছক্কার মতোই। একমাত্র পার্থক্য হলো এটাতে \(20\)-টা তল আছে।)

How many ways can you roll three $20$-sided dice such that the sum of the three rolls is exactly $42$? Here the order of the rolls matters.
(Note that a $20$-sided die is very much like a regular six-sided die other than the fact that it has $20$ faces instead of the regular six.)
"If people do not believe that mathematics is simple, it is only because they do not realize how complicated life is."
John von Neumann

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Pro_GRMR
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Re: BdMO National 2021 Higher Secondary Problem 5

Unread post by Pro_GRMR » Sun Apr 11, 2021 10:24 pm

Anindya Biswas wrote:
Sun Apr 11, 2021 9:33 pm
How many ways can you roll three $20$-sided dice such that the sum of the three rolls is exactly $42$? Here the order of the rolls matters.
(Note that a $20$-sided die is very much like a regular six-sided die other than the fact that it has $20$ faces instead of the regular six.)
Re-stating the question we get how many ways are there to write 42 as the sum of three numbers which are $\leq20$.
We do casework for the first number and find that the number cannot be $1$, as $1+20+20$ is still $41$ which is less than 42.
We also find there is only $1$ case for $2$ which is $2+20+20$. Similarly $2$ cases for $3$ and we soon realize $n-1$ cases for $n$.
Summing them up we get $1+2+\dots+19=\frac{19\cdot20}{2}=\boxed{190}$
"When you change the way you look at things, the things you look at change." - Max Planck

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