Difference between revisions of "2016 USAMO Problems/Problem 2"
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− | Define <math>v_p(N)</math> for all rational numbers <math>N</math> and primes <math>p</math>, where if <math>N=\frac{x}{y}</math>, then <math>v_p(N)=v_p(x)-v_p(y)</math>, and <math>v_p(x)</math> is the greatest power of <math>p</math> that divides <math>x</math>. Note that the expression is clearly rational, call it <math>N</math>. | + | Define <math>v_p(N)</math> for all rational numbers <math>N</math> and primes <math>p</math>, where if <math>N=\frac{x}{y}</math>, then <math>v_p(N)=v_p(x)-v_p(y)</math>, and <math>v_p(x)</math> is the greatest power of <math>p</math> that divides <math>x</math> for integer <math>x</math>. Note that the expression(that we're trying to prove is an integer) is clearly rational, call it <math>N</math>. |
<math>v_p(N)=\sum_{i=1}^\infty \lfloor \frac{k^{2}}{p^{i}} \rfloor+\sum_{j=0}^{k-1} \sum_{i=1}^\infty \lfloor \frac{j}{p^{i}}\rfloor-\sum_{j=k}^{2k-1} \sum_{i=1}^\infty \lfloor \frac{j}{p^{i}} \rfloor</math>, by Legendre. Clearly, <math>\lfloor{\frac{x}{p}}\rfloor={\frac{x-r(x,p)}{p}}</math>, and <math>\sum_{i=0}^{k-1} r(i,m)\leq \sum_{i=k}^{2n-1} r(i,m)</math>, where <math>r(i,m)</math> is the remainder function. Thus, <math>v_p(N)=\sum_{m=p^{i}, i\in \mathbb{N}}-\frac{k^{2}}{m}+\lfloor{\frac{k^{2}}{m}}\rfloor-\frac{\sum_{i=0}^{k-1} r(i,m)-\sum_{i=k}^{2k-1} r(i,m)}{m} \geq \sum_{m=p^{i}, i\in \mathbb{N}} \lceil -\frac{k^{2}}{m}+\lfloor{\frac{k^{2}}{m}}\rfloor\rceil \geq 0</math>, as the term in each summand is a sum of floors also and is clearly an integer. | <math>v_p(N)=\sum_{i=1}^\infty \lfloor \frac{k^{2}}{p^{i}} \rfloor+\sum_{j=0}^{k-1} \sum_{i=1}^\infty \lfloor \frac{j}{p^{i}}\rfloor-\sum_{j=k}^{2k-1} \sum_{i=1}^\infty \lfloor \frac{j}{p^{i}} \rfloor</math>, by Legendre. Clearly, <math>\lfloor{\frac{x}{p}}\rfloor={\frac{x-r(x,p)}{p}}</math>, and <math>\sum_{i=0}^{k-1} r(i,m)\leq \sum_{i=k}^{2n-1} r(i,m)</math>, where <math>r(i,m)</math> is the remainder function. Thus, <math>v_p(N)=\sum_{m=p^{i}, i\in \mathbb{N}}-\frac{k^{2}}{m}+\lfloor{\frac{k^{2}}{m}}\rfloor-\frac{\sum_{i=0}^{k-1} r(i,m)-\sum_{i=k}^{2k-1} r(i,m)}{m} \geq \sum_{m=p^{i}, i\in \mathbb{N}} \lceil -\frac{k^{2}}{m}+\lfloor{\frac{k^{2}}{m}}\rfloor\rceil \geq 0</math>, as the term in each summand is a sum of floors also and is clearly an integer. |
Revision as of 18:05, 28 April 2016
Problem
Prove that for any positive integer is an integer.
Solution
Define for all rational numbers and primes , where if , then , and is the greatest power of that divides for integer . Note that the expression(that we're trying to prove is an integer) is clearly rational, call it .
, by Legendre. Clearly, , and , where is the remainder function. Thus, , as the term in each summand is a sum of floors also and is clearly an integer.
See also
2016 USAMO (Problems • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAMO Problems and Solutions |