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Difference between revisions of "2005 USAMO Problems/Problem 6"

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==Problem==
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== Problem ==
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For <math>m</math> a positive integer, let <math>s(m)</math> be the sum of the digits of <math>m</math>. For <math>n\ge 2</math>, let <math>f(n)</math> be the minimal <math>k</math> for which there  exists a set <math>S</math> of <math>n</math> positive integers such that  <math>s\left(\sum_{x\in X} x\right) = k</math> for any nonempty subset <math>X\subset S</math>.  Prove that there are constants <math>0 < C_1 < C_2</math> with
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(''Titu Andreescu, Gabriel Dospinescu'') For <math>m</math> a positive integer, let <math>s(m)</math> be the sum of the digits of <math>m</math>. For <math>n\ge 2</math>, let <math>f(n)</math> be the minimal <math>k</math> for which there  exists a set <math>S</math> of <math>n</math> positive integers such that  <math>s\left(\sum_{x\in X} x\right) = k</math> for any nonempty subset <math>X\subset S</math>.  Prove that there are constants <math>0 < C_1 < C_2</math> with
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<cmath>
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<cmath>C_1 \log_{10} n \le f(n) \le C_2 \log_{10} n.</cmath>
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C_1 \log_{10} n \le f(n) \le C_2 \log_{10} n.
 
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</cmath>
 
  
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==Solution==
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== Solution ==
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{{solution}}
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For the upper bound, let <math>p</math> be the smallest integer such that <math>10^p\geq n(n+1)/2</math> and let
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By Jason's theorem number 1239142, we know this is true.
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<cmath>S = \{10^p - 1, 2(10^p - 1), \ldots, n(10^p - 1)\}.</cmath>
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The sum of any nonempty set of elements of <math>S</math> will have the form <math>k(10^p - 1)</math> for some <math>1\leq k\leq n(n+1)/2</math>. Write <math>k(10^p - 1) = [(k - 1)10^p] + [(10^p - 1) - (k - 1)]</math>. The second term gives the bottom <math>p</math> digits of the sum and the first term gives at most <math>p</math> top digits. Since the sum of a digit of the second term and the corresponding digit of <math>k - 1</math> is always 9, the sum of the digits will be <math>9p</math>. Since <math>10^{p-1} < n(n+1)/2</math>, this example shows that
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<cmath>f(n)\leq 9p\leq 9\log_{10}(5n(n+1)).</cmath>
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Since <math>n\geq 2</math>, <math>5(n + 1) < n^4</math>, and hence
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<cmath>f(n) < 9\log_{10}n^5 = 45\log_{10}n.</cmath>
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For the lower bound, let <math>S</math> be a set of <math>n\geq 2</math> positive integers such that any nonempty <math>X\subset S</math> has <math>s\left(\sum_{x\in X} x\right) = f(n)</math>. Since <math>s(m)</math> is always congruent to <math>m</math> modulo 9, <math>\sum_{x\in X} x\equiv f(n)\pmod{9}</math> for all nonempty <math>X\subset S</math>. Hence every element of <math>S</math> must be a multiple of 9 and <math>f(n)\geq 9</math>. Let <math>q</math> be the largest positive integer such that <math>10^q - 1\leq n</math>. Lemma 1 below shows that there is a nonempty subset <math>X</math> of <math>S</math> with <math>\sum_{x\in X} x</math> a multiple of <math>10^q - 1</math>, and hence Lemma 2 shows that <math>f(n)\geq 9q</math>.
  
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== See Also==
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'''Lemma 1.''' Any set of <math>m</math> positive integers contains a nonempty subset whose sum is a multiple of <math>m</math>.
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''Proof.'' Suppose a set <math>T</math> has no nonempty subset with sum divisible by <math>m</math>. Look at the possible sums mod <math>m</math> of nonempty subsets of <math>T</math>. Adding a new element <math>a</math> to <math>T</math> will give at least one new sum mod <math>m</math>, namely the least multiple of <math>a</math> which does not already occur. Therefore the set <math>T</math> has at least <math>|T|</math> distinct sums mod <math>m</math> of nonempty subsets and <math>|T| < m</math>. <math>\blacksquare</math>
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'''Lemma 2.''' Any positive multiple <math>M</math> of <math>10^q - 1</math> has <math>s(M)\geq 9q</math>.
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''Proof.'' Suppose on the contrary that <math>M</math> is the smallest positive multiple of <math>10^q - 1</math> with <math>s(M) < 9q</math>. Then <math>M\neq 10^q - 1</math>, hence <math>M > 10^q</math>. Suppose the most significant digit of <math>M</math> is the <math>10^m</math> digit, <math>m\geq q</math>. Then <math>N = M - 10^{m-q}(10^q - 1)</math> is a smaller positive multiple of <math>10^q - 1</math> and has <math>s(N)\leq s(M) < 9q</math>, a contradiction. <math>\blacksquare</math>
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Finally, since <math>10^{q+1} > n</math>, we have <math>q + 1 > \log_{10} n</math>. Since <math>f(n)\geq 9q</math> and <math>f(n)\geq 9</math>, we have
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<cmath>f(n)\geq \frac{9q + 9}{2} > \frac{9}{2}\log_{10}n.</cmath>
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Weaker versions of Lemmas 1 and 2 are still sufficient to prove the desired type of lower bound.
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{{alternate solutions}}
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== See Also ==
 
{{USAMO newbox|year=2005|num-b=5|after=Last Question}}
 
{{USAMO newbox|year=2005|num-b=5|after=Last Question}}
 
{{MAA Notice}}
 
{{MAA Notice}}
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sadfsdfafasdfsadfsadfasdf
 

Latest revision as of 22:59, 29 March 2016

Problem

(Titu Andreescu, Gabriel Dospinescu) For $m$ a positive integer, let $s(m)$ be the sum of the digits of $m$. For $n\ge 2$, let $f(n)$ be the minimal $k$ for which there exists a set $S$ of $n$ positive integers such that $s\left(\sum_{x\in X} x\right) = k$ for any nonempty subset $X\subset S$. Prove that there are constants $0 < C_1 < C_2$ with \[C_1 \log_{10} n \le f(n) \le C_2 \log_{10} n.\]

Solution

For the upper bound, let $p$ be the smallest integer such that $10^p\geq n(n+1)/2$ and let \[S = \{10^p - 1, 2(10^p - 1), \ldots, n(10^p - 1)\}.\] The sum of any nonempty set of elements of $S$ will have the form $k(10^p - 1)$ for some $1\leq k\leq n(n+1)/2$. Write $k(10^p - 1) = [(k - 1)10^p] + [(10^p - 1) - (k - 1)]$. The second term gives the bottom $p$ digits of the sum and the first term gives at most $p$ top digits. Since the sum of a digit of the second term and the corresponding digit of $k - 1$ is always 9, the sum of the digits will be $9p$. Since $10^{p-1} < n(n+1)/2$, this example shows that \[f(n)\leq 9p\leq 9\log_{10}(5n(n+1)).\] Since $n\geq 2$, $5(n + 1) < n^4$, and hence \[f(n) < 9\log_{10}n^5 = 45\log_{10}n.\] For the lower bound, let $S$ be a set of $n\geq 2$ positive integers such that any nonempty $X\subset S$ has $s\left(\sum_{x\in X} x\right) = f(n)$. Since $s(m)$ is always congruent to $m$ modulo 9, $\sum_{x\in X} x\equiv f(n)\pmod{9}$ for all nonempty $X\subset S$. Hence every element of $S$ must be a multiple of 9 and $f(n)\geq 9$. Let $q$ be the largest positive integer such that $10^q - 1\leq n$. Lemma 1 below shows that there is a nonempty subset $X$ of $S$ with $\sum_{x\in X} x$ a multiple of $10^q - 1$, and hence Lemma 2 shows that $f(n)\geq 9q$.

Lemma 1. Any set of $m$ positive integers contains a nonempty subset whose sum is a multiple of $m$.

Proof. Suppose a set $T$ has no nonempty subset with sum divisible by $m$. Look at the possible sums mod $m$ of nonempty subsets of $T$. Adding a new element $a$ to $T$ will give at least one new sum mod $m$, namely the least multiple of $a$ which does not already occur. Therefore the set $T$ has at least $|T|$ distinct sums mod $m$ of nonempty subsets and $|T| < m$. $\blacksquare$

Lemma 2. Any positive multiple $M$ of $10^q - 1$ has $s(M)\geq 9q$.

Proof. Suppose on the contrary that $M$ is the smallest positive multiple of $10^q - 1$ with $s(M) < 9q$. Then $M\neq 10^q - 1$, hence $M > 10^q$. Suppose the most significant digit of $M$ is the $10^m$ digit, $m\geq q$. Then $N = M - 10^{m-q}(10^q - 1)$ is a smaller positive multiple of $10^q - 1$ and has $s(N)\leq s(M) < 9q$, a contradiction. $\blacksquare$

Finally, since $10^{q+1} > n$, we have $q + 1 > \log_{10} n$. Since $f(n)\geq 9q$ and $f(n)\geq 9$, we have \[f(n)\geq \frac{9q + 9}{2} > \frac{9}{2}\log_{10}n.\] Weaker versions of Lemmas 1 and 2 are still sufficient to prove the desired type of lower bound.

Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

See Also

2005 USAMO (Problems • Resources)
Preceded by
Problem 5 		Followed by
Last Question
1 • 2 • 3 • 4 • 5 • 6
All USAMO Problems and Solutions

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