Below, we present a problem from the 1/28 Algebra and Number Theory set, along with 3 student submitted solution. You may vote up for the solutions that you think should be featured, and should vote down for those solutions that you think are wrong (voting is anonymous!). Also, feel free to make remarks about these solutions, especially since threading of comments has been introduced :).

Minimum Digit Sum Fraction Let \(S(N) \) denote the digit sum of the integer \(N\). As \(N\) ranges over all 3-digit positive numbers, what value of \(N\) would give the minimum of \(M = \frac{N}{S(N)}\)?

You may try the problem by clicking on the above link.

All solutions may have LaTeX edits to make the math appear properly. The exposition is presented as is, and has not been edited.

\[ \mbox{Remarks from Calvin} \]

Solution A - This has been edited to "Edited Solution A". Please note that the phrase "\(A\) is more dominant on the numerator than on the denominator" has been removed, is it's not a valid reasoning for why we want to minimize A. This logic is flawed, in part since we want to ended up maximizing B, which is also 'more dominant on the numerator'. This was used in several solutions, which were marked wrong.

Edited Solution A - This is much clearer about how to proceed. The main idea being that in dealing with a rational function, it's easier to minimize / maximize it when the variable only appears in the numerator or denominator. There are several ways that work, and we simply used long division. There are several ways we could have proceeded. A slightly more straightforward way, is to repeat the same analysis with \(A\) to obtain \( M = \frac {100A + 10 B + C}{A+B+C} = 100 - \frac {90A + 99C}{A+B+C}\), where it's obvious that we would want \(A=1\). Working with \(B\) is slightly trickier, \( M = \frac {100A + 10 B + C}{A+B+C} = 10 + \frac {90A - 9C}{A+B+C}\), the numerator is always positive since \(A\geq 1\), hence we want to maximize \(B\). [For those who have learnt calculus, this is essentially partial differentiation.]

Solution B - This solution makes no sense at all. I am unable to find any information on the 'batting average lemma', though I can guess what it should mean. However, the way it is stated, it's quite clearly false, like the counter example \( \frac {99}{200} \leq \frac {1} {2} \) but \( \frac {299}{3} > \frac {1}{2} \). I am guessing that he intended it to be \( \frac {a}{b} \leq \frac {a+c}{b+d} \leq \frac {c}{d} \), which is true for \(b, d > 0\). If you want to state a theorem, make sure that the statement is correct, and that the proper conditions are given.

Solution C - This was the most common error made, arguing that the numerator must be minimum (which occurs when \(N=100\)) and that the denominator must be maximum (which occurs when \(N=999\)). This statement is false, as the is great dependence between the numerator and denominator. The rest of the solution devolves into case checking. It is far better to have simply done a solution through case checking, and being thorough in explaining "the pattern ... increase the numerator by 10, the fraction keeps decreasing", which can be shown using Edited Solution A.

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TopNewestSolution A - This is a 3-digit number, so we can call the digits A B C, where \( A \ne 0 \). This means: \( N = 100A + 10B + C \) and \( S = A + B + C \). So, by the definition of \( M \), we have: \( M = \frac {100A + 10B + C}{A + B + C} \) Remark that this can be re-written as \( M = \frac {99A + 9B}{A + B + C} + 1 \) We want to minimize this, so we want \( C = 9 \). Substituting this into the equation above, we obtain: \( M = \frac {99A + 9B}{A+B+9} + 1 = \frac {(90A-81)(9A+9B+81)}{A+B+9} + 1 \) \( = \frac {90A-81}{A+B+9} + 9 + 1 = \frac {90A-81}{A+B+9} + 10 \) Again, we wish to minimize this so, clearly, \( B = 9 \). Substituting \( B = 9 \) into this, we get: \( M = \frac {90A-81}{A+18} + 10. \) Now, haste makes waste! We cannot just conclude that \( A = 9 \) as well. In fact, it's quite clear that 999 isn't the best we can do. Note that, unlike B and C, A is more dominant on the numerator than on the denominator. We do not have to long-divide to see this. In fact, I am skipping that detail completely in my solution because it's seen quite easily. So, to minimize the expression, we need to minimize A, thus, it needs to be 1. We got: A = 1, B = 9, C = 9, so the answer is \( \fbox{199} \). – Calvin Lin Staff · 4 years, 6 months ago

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– Ahaan Rungta · 4 years, 6 months ago

Oh, this is my solution! Calculus isn't required. I actually realized, right after I clicked "Submit Solution" that I could have made the A-minimization a little clearer. But I guess it looks clear! Thanks to those who voted for this. Currently leading. :DLog in to reply

– Calvin Lin Staff · 4 years, 6 months ago

Actually, the entire thing could be made a lot clearer. Can you email me an updated version?Log in to reply

– Ahaan Rungta · 4 years, 6 months ago

Sure! What is your e-Mail address? Actually, I would only update the last section (the A-minimization part). But maybe that part would make it clearer. I'll e-Mail it to you once I have your e-Mail address.Log in to reply

– Tim Ye · 4 years, 6 months ago

calvin@brilliant.orgLog in to reply

– Calvin Lin Staff · 4 years, 6 months ago

Lol! Thanks. I emailed him directly so that I can send him his proof, instead of having him write up the latex portions again.Log in to reply

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– Calvin Lin Staff · 4 years, 6 months ago

Do you need quadratic / calculus? Try what the author suggested, it's not hard if you keep to the approaches used in the first 2 steps.Log in to reply

Edited Solution A - This is a 3-digit number, so we can call the digits A B C, where \( A \ne 0 \). This means: \( N = 100A + 10B + C \) and \( S = A + B + C \). So, by the definition of \( M \), we have: \( M = \frac {100A + 10B + C}{A + B + C} \)

Remark that this can be re-written as \[ M = \frac {99A + 9B}{A + B + C} + 1 \]. Consider this as a function of C and notice that C only appears in the denominator. We want to minimize this, so we want to maximize C; i.e. \( C = 9 \). [Note: It doesn't matter what positive values A and B are. - Calvin]

Substituting this into the equation above, we obtain: \( M = \frac {99A + 9B}{A+B+9} + 1 = \frac {(90A-81)(9A+9B+81)}{A+B+9} + 1 \). Now, \(M\) can be rewritten as \[ M = \frac {90A-81}{A+B+9} + 9 + 1 = \frac {90A-81}{A+B+9} + 10.\] Again, we wish to minimize this so, in the same manner, we note that B only appears in the denominator in this function of B. Clearly, we want to maximize B to minimize this fraction. So, \( B = 9 \).

Substituting \( B = 9 \) into this, we get: \( M = \frac {90A-81}{A+18} + 10. \) Now, haste makes waste! We cannot just conclude that \( A = 9 \) as well. In fact, it's quite clear that 999 isn't the best we can do. Note that \[ M = \frac {90A-81}{A + 18} + 10 = \frac {90A + 90 \times 18 - 90 \times 18 - 81} { A + 18} + 10 = 90 - \frac {90 \times 18 + 81} { A + 18} + 10 . \] Aha! We want to minimize this expression, so we want to maximize \( \frac {1}{A+18} \). And, to do this, we need to minimize A, since A is in the denominator. This is reverse logic from what we did with B and C. Since A can't be 0 (if it was, we wouldn't have a 3-digit number), A must be 1.

So, we got: A = 1, B = 9, C = 9, so the answer is \( \fbox{199} \). – Calvin Lin Staff · 4 years, 6 months ago

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– Sayan Chaudhuri · 4 years, 6 months ago

calvin u r really a mathematics-charmer!!!HATS OFF FOR ENDLESS EFFORT FOR CLARIFYING .......Log in to reply

– Jayver De Torres · 4 years, 6 months ago

How is $/frac{99A + B}{A+B+9} +1 = /frac{(90A-81)(9A+B + 81)}{A+B+9} + 1$ happen?Log in to reply

– Ahaan Rungta · 4 years, 6 months ago

Thank you for posting the new updated solution! So the number of votes would be the number of votes for the previous one + this one?Log in to reply

But I think it is the one which will be featured. – Zi Song Yeoh · 4 years, 6 months ago

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[James suggested a problem, which I might use in an upcoming set - Calvin] – James Lin · 4 years, 6 months ago

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– Calvin Lin Staff · 4 years, 6 months ago

THat's an interesting question. Would you like to send your solution to me? I'd like to use it as an upcoming problem.Log in to reply

– James Lin · 4 years, 6 months ago

Oh, I have a somewhat ugly solution; I was looking to see if anyone had a good solution to it also.Log in to reply

[Solution edited out - Calvin] – Aakash Kansal · 4 years, 6 months ago

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Remarks Added. – Calvin Lin Staff · 4 years, 6 months ago

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It is a very simple one. It is very obvious that to make the fraction smallest, we will have to make the number's hundredth digit small i.e. 1. Now, it is again very obvious that its units digit should be 9 so that the fraction becomes small (as denominator will increase same as numerator). Now, we wish to maximize the denominator and so the tenths digit must be 9. So 199 is the answer. – Brilliant Kumar · 4 years, 6 months ago

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– Calvin Lin Staff · 4 years, 6 months ago

Solutions like yours have been marked wrong. You offer no explanation of why the value will be minimized, since the numerator and denominator both change. See Solution C.Log in to reply

– Brilliant Kumar · 4 years, 6 months ago

I have an explanation. It is really very good but it will take a lot of formatting to show it here. Please send me your email id so that I can mail a document which has the explanation. Please give me a chance to explain my answer. Please sir.Log in to reply

– Calvin Lin Staff · 4 years, 6 months ago

You can email solutions to mathematics@brilliant.org. If you intend to send a scanned hand-written copy, please ensure that it is legible. Thanks.Log in to reply

– Brilliant Kumar · 4 years, 6 months ago

I have mailed the solution. View it once.Log in to reply

– Brilliant Kumar · 4 years, 6 months ago

Actually I am quite busy these days. I'll send the solution in a day or two. Is it ok?Log in to reply

No simple solution. :( – Zi Song Yeoh · 4 years, 6 months ago

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– Calvin Lin Staff · 4 years, 6 months ago

The solution is actually really simple and direct, based off knowing the graph of \( \frac {1}{X}\).Log in to reply

– Zi Song Yeoh · 4 years, 6 months ago

Which solution?Log in to reply

– Ahaan Rungta · 4 years, 6 months ago

Both Solutions A and C show this, but I just can't deny that mine is a little clearer on that. :PLog in to reply

CANT GET B ! C IS JUST USELESS and A IS QUITE CORRECT AND SHORT – Aakash Kansal · 4 years, 6 months ago

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Solution B - Letting our three digit number \(def\) be such that \(def = 100a +10b +c\), we see that \(M=\frac{100a+10b+c}{a+b+c}\). The batting average lemma states that if \(\frac{a}{b} \le \frac{c}{d}\) then \(\frac{a+b}{c+d} \le \frac{c}{d}\). So now we start with \(n=100\) and consider what happens if we add, batting average style, to each digits place: we add a fraction of value corresponding to the value of the place. So, and this is not fully rigorous but it does suffice after a little numerical testing at the extreme ranges to see if the pattern holds, we see that increasing \(b\) and \(c\) decreases the value of \(M\) but that increasing \(a\) does not. So we let \(b\) and \(c\) be \(9\) and \(a\) be \(1\) and our answer is that the minimum value of \(M\) occurs at \(n=\boxed{199}\) – Calvin Lin Staff · 4 years, 6 months ago

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Solution C - Here, the numerator (N) of the fraction should be minimum and denominator (S(N)) should be maximum. Hence, the numerator must be in 100's and denominator should be an addition of the digits of max. magnitude (i.e, 9). Summing up, we get, Numerator can be 109, 119,... and Denominator can be (1+0+9), (1+1+9),... respectively. By trial and error,

109/(1+0+9)= 10.9; 119/(1+1+9)= 10.81; and so on.... Hence, observing the pattern we see that as we increase the numerator by 10, the fraction keeps decreasing. Hence at199/(1+9+9)= 10.47, the fraction should be minimum. For further verifying it, we take299/(2+9+9)= 14.95;399/(3+9+9)= 19; and so the value of the fraction keeps increasing as we increase the numerator. Hence, we conclude that the fraction "M" has minimum value when N=199. – Calvin Lin Staff · 4 years, 6 months agoLog in to reply

– Sri Venkata Vivek Dhulipala · 4 years, 6 months ago

post each and every solution.They are goodLog in to reply