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# 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
3 years, 9 months ago

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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. · 3 years, 9 months ago

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? · 3 years, 9 months ago

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. · 3 years, 9 months ago

Ohh.. I see.. brilliant! Thanks :) · 3 years, 9 months ago

It is pretty easy to see why there must be at least 79 or more '1's in this number. A really interesting problem. · 3 years, 9 months ago

Could you show us your proof of this fact? · 3 years, 9 months ago

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 · 3 years, 9 months ago

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... · 3 years, 9 months ago

8322111111111111111111111111111111111111111111111111111111111111111111111111111111111

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

8322 with 81 1s at the back · 3 years, 9 months ago

Could you show us your proof of your answer Rohan?? ^^ · 3 years, 9 months ago

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! · 3 years, 9 months ago

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. · 3 years, 9 months ago