Day 8: Eight Numbers in a Hat
A teacher organises a game to improve her students' arithmetic.
She puts the numbers 0 , 1 , 2 , … , 7 into a hat.
Every turn one student takes out two numbers from the hat and replaces them with the sum of their sum and their product (so the number of numbers in the hat decreases by one).
The game ends when there is one number left in the hat. Incredibly, it turns out that this number is always the same. What is this number?
This problem is part of the Advent Calendar 2015 .
The answer is 40319.
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1 solution
8 ! − 1 = 4 0 3 1 9
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Hmm thank you! I must have been very tired... Is there any way to credit those who got it right and change the answer?
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@Brilliant Mathematics , pls fix this thanks
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@Pi Han Goh – Yes I have edited my solution; sorry and thank you again!
Yes Same Way.
Your solution is not inclusive of cases where we apply the sum-product function on pairs such as, say, something like ( a + 1 ) ( b + 1 ) − 1 and ( c + 1 ) ( d + 1 ) − 1
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Ah I did think of that; I have included that by saying at any step (in bold). So at any step we will add one to the expression and multiply, then take away one. Sorry for it not being very clear; I wasn't very sure how to phrase it in the best way.
Alternatively, you could note that a ∗ b : = a + b + a b is both commutative and associative, just like addition and multiplication.
Yours is much cleverer, though.
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Ah yes you are very correct; nice observation! In fact, that was my initial idea for a solution but then I wanted to make it more explicit; I think all I have done is (informally) show why the operation is associative (and commutative). Thank you :)
You can do this once and simply type in your final answer, but why is it always the same answer?
At each step you replace a and b with a + b + a b = ( a + 1 ) ( b + 1 ) − 1 .
Now note that if we have some c that we use with this number at any step , then we will get the number ( ( a + 1 ) ( b + 1 ) − 1 + 1 ) ( c + 1 ) − 1 = ( a + 1 ) ( b + 1 ) ( c + 1 ) − 1 by applying the factorisation above. Continuing this idea we find that no matter what, we will end with the number ( a + 1 ) ( b + 1 ) … ( g + 1 ) − 1 Substituting in our numbers gives 8 ! − 1 = 4 0 3 1 9 as required.