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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

@A Former 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...

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
–
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.

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

Let the digits be $a_{85}\geq a_{84}\geq \cdots \geq a_1$ such that $a_1+a_2+\cdots+a_{85}=a_1a_2\cdots a_{85}$, then $\frac{a_1}{a_1a_2\cdots a_{85}}+\frac{a_2}{a_1a_2\cdots a_{85}}+\cdots+\frac{a_{85}}{a_1a_2\cdots a_{85}}=1.$ we have 85 fractions and the greatest of them is the last one, therefore $\frac{a_{85}}{a_1a_2\cdots a_{85}}\geq \frac{1}{85}$ and then $a_1a_2\cdots a_{84}\leq85$.

At most six digits among $a_1,a_2,\ldots, a_{84}$ are greater than 1, therefore, at least 78 digits are equal to 1.

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 $85$ with highest number of divisors with one digit and I think according to your argument such integer has at most $6$ one digit divisors. Right?

Each of $a_1, a_2, \ldots, a_{84}$ is a positive integer (zeros are not allowed). Suppose seven digits among $a_1, a_2, \ldots, a_{84}$ are greater than 1, i.e. are $\geq2$, then the product $a_1a_2\cdots a_{84}$ will be $\geq 2^7=128$ which is a contradiction. Therefore, at most six digits among $a_1, a_2, \ldots, a_{84}$ can be greater than 1, thus, at least 78 of them are equal to 1.

Easy Math Editor

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## Comments

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

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Could you show us your proof of this fact?

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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 2

22..*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'sLog in to reply

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8322111111111111111111111111111111111111111111111111111111111111111111111111111111111

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

8322 with 81 1s at the back

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

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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!

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

Let the digits be $a_{85}\geq a_{84}\geq \cdots \geq a_1$ such that $a_1+a_2+\cdots+a_{85}=a_1a_2\cdots a_{85}$, then $\frac{a_1}{a_1a_2\cdots a_{85}}+\frac{a_2}{a_1a_2\cdots a_{85}}+\cdots+\frac{a_{85}}{a_1a_2\cdots a_{85}}=1.$ we have 85 fractions and the greatest of them is the last one, therefore $\frac{a_{85}}{a_1a_2\cdots a_{85}}\geq \frac{1}{85}$ and then $a_1a_2\cdots a_{84}\leq85$.

At most six digits among $a_1,a_2,\ldots, a_{84}$ are greater than 1, therefore, at least 78 digits are equal to 1.

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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 $85$ with highest number of divisors with one digit and I think according to your argument such integer has at most $6$ one digit divisors. Right?

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Each of $a_1, a_2, \ldots, a_{84}$ is a positive integer (zeros are not allowed). Suppose seven digits among $a_1, a_2, \ldots, a_{84}$ are greater than 1, i.e. are $\geq2$, then the product $a_1a_2\cdots a_{84}$ will be $\geq 2^7=128$ which is a contradiction. Therefore, at most six digits among $a_1, a_2, \ldots, a_{84}$ can be greater than 1, thus, at least 78 of them are equal to 1.

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