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Is an easy way to solve the problem?

The product of the digits of positive integer $n$ is $20$, and the sum of the digits is $13$. What is the smallest possible value of $n$?

The way I did is to list equations $$n_1+n_2+n_3+...+n_i=13$$ $$n_1*n_2*n_3*...*n_i=20$$ Then I tried $n_s$. I don't think this is not a very good method.

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  • $\begingroup$ First off, you wrote your equations backwards. The sum is 13 and the product is 20. $\endgroup$ – Oscar Lanzi Jun 11 '16 at 17:36
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First, let's list out all of the ways $20$ can be factored with digits: $$20=2*2*5$$ $$20=4*5$$

If we take $225$, we can add four $1$s at the beginning to get a number with a sum of $13$, so we have $1111225$.

If we take $45$, we can add four $1$s at the beginning to get a number with a sum of $13$, so we have $111145$.

Clearly, $111145$ is the smaller number, so that is our answer.

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We see that $20$ can be factored as $2\times2\times5$.So,we can use only these three numbers or add $1$'s to our number since multiplying by $1$ gives same result.

So,we see,13 can be written as $2+2+5+1+1+1+1$.We add $4$ ones without changing the product.

So,least number possible with this is $111145$.

Note that we multiply the two $2$'s to form $4$ to give the least possible number.

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  • $\begingroup$ Beat me to it +1. This is what I fet for working from a cell phone. Still would have been an awesome pinball score when I was younger. $\endgroup$ – Oscar Lanzi Jun 11 '16 at 17:34
  • $\begingroup$ @OscarLanzi I don't quite really understand what you are saying....can you please be more explanatory?:-) $\endgroup$ – tatan Jun 11 '16 at 17:36
  • $\begingroup$ You beat me to the answer. I was too slow, hard to type from a cell phone without errors. $\endgroup$ – Oscar Lanzi Jun 11 '16 at 17:38
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    $\begingroup$ @OscarLanzi Ok....I understand the first part of your comment....now what do you mean by the second part-"Still would have been an awesome pinball score when I was younger." ? $\endgroup$ – tatan Jun 11 '16 at 17:40
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Hint: If the product is $20=2 \cdot 2 \cdot 5 = 4 \cdot 5$, then the only possible digits are $1,2,4,5$.

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Since $n_1,\dots,n_i$ are positive integers and $20$ is primarily decomposed as $20=2^2\cdot 5$, we quickly find the largest $n$, that is $5221111$. But before finding the smallest, lets note since $20$ is a multiple of $5$ and no digits other than $0$ and $5$ aren't divisible by $5$. and since product of the digits is non-zero, one and only one of the digits is $5$, but by similar but not not same reasoning we can say either, one and only one digit is $4$, or, two and exactly two digits are $2$. in either ways the some of non-units is $9$. So, we have 4 other digits, each of which are unit. so the numbers we can write with the properties asserted, have either, 6, or 7 digits. of course the smallest among them should have 6 digit. and is $111145$

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  • $\begingroup$ The sum of the digits is $13$, not the product. Also, I think we assume base-$10$, so there's no way $13$ can be a digit. $\endgroup$ – Noble Mushtak Jun 11 '16 at 17:40
  • $\begingroup$ Yeah... I first thought it is a hard problem... ;) $\endgroup$ – Omid Ghayour Jun 11 '16 at 18:06

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