Help,its so hard for me

Determine the biggest integers 85-digit number which satisfies the sum of all digits equal to the multiplication of all numbers

Note by Valian Fil Ahli
5 years, 11 months ago

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16 votes

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It is pretty easy to see why there must be at least 79 or more '1's in this number. A really interesting problem.

Brilliant Member - 5 years, 10 months ago

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Could you show us your proof of this fact?

Jorge Tipe - 5 years, 10 months ago

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Let there be less than or equal to 78 '1's then the product of the remaining digits is at least 222..*2 (product of 7 twos) which is 128. But this number(222222211.....1) does not satisfy the given condition. Replacing any '2' which any other digit still won't make the number satisfy the given condition( since the number will just get larger) nor will replacing a '1' by any other digit.( since the product gets larger). We want a smaller product. Hence there should be at least 79 '1's

Brilliant Member - 5 years, 10 months ago

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@Brilliant Member Actually my proof is the same as your proof above mine but you say that at most there are 6 digit not equal to one and so there must be 85 - 6 = 79, at least 79 '1's...

Brilliant Member - 5 years, 10 months ago

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8322111111111111111111111111111111111111111111111111111111111111111111111111111111111

highest 1 I got............ only 1 I got

8322 with 81 1s at the back

Rohan Subagaran - 5 years, 11 months ago

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Could you show us your proof of your answer Rohan?? ^^

Valian Fil Ahli - 5 years, 10 months ago

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8x3x2x2=96.Since the other numbers are 1s, when they are multiplied by any number, it equals to the number you multiplied it by. 8+3+2+2=15, there are 81 1s that add up to 81. 81+15 =96. I hope this helped!

Rohan Subagaran - 5 years, 10 months ago

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@Rohan Subagaran I think he means to say ''prove that this is the largest number and there is no other larger number''. I think a simple case by case analysis( of a lot of cases) should suffice.

Brilliant Member - 5 years, 10 months ago

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@Brilliant Member I said it was the highest I got, not the answer

Rohan Subagaran - 5 years, 10 months ago

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I'm not sure about the answer, but I think this will help you.

Let the digits be a85a84a1a_{85}\geq a_{84}\geq \cdots \geq a_1 such that a1+a2++a85=a1a2a85a_1+a_2+\cdots+a_{85}=a_1a_2\cdots a_{85}, then a1a1a2a85+a2a1a2a85++a85a1a2a85=1.\frac{a_1}{a_1a_2\cdots a_{85}}+\frac{a_2}{a_1a_2\cdots a_{85}}+\cdots+\frac{a_{85}}{a_1a_2\cdots a_{85}}=1. we have 85 fractions and the greatest of them is the last one, therefore a85a1a2a85185\frac{a_{85}}{a_1a_2\cdots a_{85}}\geq \frac{1}{85} and then a1a2a8485a_1a_2\cdots a_{84}\leq85.

At most six digits among a1,a2,,a84a_1,a_2,\ldots, a_{84} are greater than 1, therefore, at least 78 digits are equal to 1.

Jorge Tipe - 5 years, 11 months ago

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Can you please explain the last sentence in your argument? It is not very obvious to me. I think we need to find the integer less than or equal to 8585 with highest number of divisors with one digit and I think according to your argument such integer has at most 66 one digit divisors. Right?

Snehal Shekatkar - 5 years, 11 months ago

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Each of a1,a2,,a84a_1, a_2, \ldots, a_{84} is a positive integer (zeros are not allowed). Suppose seven digits among a1,a2,,a84a_1, a_2, \ldots, a_{84} are greater than 1, i.e. are 2\geq2, then the product a1a2a84a_1a_2\cdots a_{84} will be 27=128\geq 2^7=128 which is a contradiction. Therefore, at most six digits among a1,a2,,a84a_1, a_2, \ldots, a_{84} can be greater than 1, thus, at least 78 of them are equal to 1.

Jorge Tipe - 5 years, 11 months ago

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@Jorge Tipe Ohh.. I see.. brilliant! Thanks :)

Snehal Shekatkar - 5 years, 10 months ago

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