https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Hrithikguy&feedformat=atomAoPS Wiki - User contributions [en]2024-03-29T13:58:42ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2011_USAJMO_Problems/Problem_1&diff=382632011 USAJMO Problems/Problem 12011-04-29T00:43:53Z<p>Hrithikguy: </p>
<hr />
<div>Find, with proof, all positive integers <math>n</math> for which <math>2^n + 12^n + 2011^n</math> is a perfect square.<br />
<br />
<br />
<br />
Let <math>2^n + 12^n + 2011^n = x^2</math><br />
<math>(-1)^n + 1 \equiv x^2 \pmod {3}</math>.<br />
Since all perfect squares are congruent to 0 or 1 modulo 3, this means that n must be odd.<br />
Proof by Contradiction:<br />
I will show that the only value of <math>n</math> that satisfies is <math>n = 1</math>.<br />
Assume that <math>n \ge 2</math>. <br />
Then consider the equation<br />
<math>2^n + 12^n = x^2 - 2011^n</math>.<br />
From modulo 2, we easily that x is odd. Let <math>x = 2a + 1</math>, where a is an integer.<br />
<math>2^n + 12^n = 4a^2 + 4a + 1 - 2011^n</math>.<br />
Dividing by 4,<br />
<math>2^{n-2} + 3^n \cdot 4^{n-1} = a^2 + a + \dfrac {1}{4} (1 - 2011^n})</math>.<br />
Since <math>n \ge 2</math>, <math>n-2 \ge 0</math>, so <math>2^{n-2}</math> similarly, the entire LHS is an integer, and so are <math>a^2</math> and <math>a</math>. Thus, <math> \dfrac {1}{4} (1 - 2011^n})</math> must be an integer.<br />
Let <math> \dfrac {1}{4} (1 - 2011^n}) = k</math>. Then we have <math>1- 2011^n = 4k</math>.<br />
<math>1- 2011^n \equiv 0 \pmod {4}</math><br />
<math>(-1)^n \equiv 1 \pmod {4}</math>.<br />
Thus, n is even.<br />
However, I have already shown that <math>n</math> must be odd. This is a contradiction. Therefore, <math>n</math> is not greater than or equal to 2, and must hence be less than 2. The only positive integer less than 2 is 1.<br />
-hrithikguy</div>Hrithikguyhttps://artofproblemsolving.com/wiki/index.php?title=2011_USAJMO_Problems/Problem_1&diff=382622011 USAJMO Problems/Problem 12011-04-29T00:39:19Z<p>Hrithikguy: Created page with 'Let <math>2^n + 12^n + 2011^n = x^2</math> <math>(-1)^n + 1 \equiv x^2 \pmod {3}</math>. Since all perfect squares are congruent to 0 or 1 modulo 3, this means that n must be odd…'</p>
<hr />
<div>Let <math>2^n + 12^n + 2011^n = x^2</math><br />
<math>(-1)^n + 1 \equiv x^2 \pmod {3}</math>.<br />
Since all perfect squares are congruent to 0 or 1 modulo 3, this means that n must be odd.<br />
Proof by Contradiction:<br />
I will show that the only value of <math>n</math> that satisfies is <math>n = 1</math>.<br />
Assume that <math>n \ge 2</math>. <br />
Then consider the equation<br />
<math>2^n + 12^n = x^2 - 2011^n</math>.<br />
From modulo 2, we easily that x is odd. Let <math>x = 2a + 1</math>, where a is an integer.<br />
<math>2^n + 12^n = 4a^2 + 4a + 1 - 2011^n</math>.<br />
Dividing by 4,<br />
<math>2^{n-2} + 3^n \cdot 4^{n-1} = a^2 + a + \dfrac {1}{4} (1 - 2011^n})</math>.<br />
Since <math>n \ge 2</math>, <math>n-2 \ge 0</math>, so <math>2^{n-2}</math> similarly, the entire LHS is an integer, and so are <math>a^2</math> and <math>a</math>. Thus, <math> \dfrac {1}{4} (1 - 2011^n})</math> must be an integer.<br />
Let <math> \dfrac {1}{4} (1 - 2011^n}) = k</math>. Then we have <math>1- 2011^n = 4k</math>.<br />
<math>1- 2011^n \equiv 0 \pmod {4}</math><br />
<math>(-1)^n \equiv 1 \pmod {4}</math>.<br />
Thus, n is even.<br />
However, I have already shown that <math>n</math> must be odd. This is a contradiction. Therefore, <math>n</math> is not greater than or equal to 2, and must hence be less than 2. The only positive integer less than 2 is 1.<br />
-hrithikguy</div>Hrithikguyhttps://artofproblemsolving.com/wiki/index.php?title=1985_AJHSME_Problems/Problem_2&diff=345921985 AJHSME Problems/Problem 22010-06-05T02:42:30Z<p>Hrithikguy: /* Solution 3 */</p>
<hr />
<div>==Problem==<br />
<br />
<math>90+91+92+93+94+95+96+97+98+99=</math><br />
<br />
<br />
<math>\text{(A)}\ 845 \qquad \text{(B)}\ 945 \qquad \text{(C)}\ 1005 \qquad \text{(D)}\ 1025 \qquad \text{(E)}\ 1045</math><br />
<br />
==Solution==<br />
<br />
===Solution 1===<br />
We could just add them all together. But what would be the point of doing that? So we find a slicker way.<br />
<br />
We find a simpler problem in this problem, and simplify -> <math>90 + 91 + ... + 98 + 99 = 90 \times 10 + 1 + 2 + 3 + ... + 8 + 9</math><br />
<br />
We know <math>90 \times 10</math>, that's easy - <math>900</math>. So how do we find <math>1 + 2 + ... + 8 + 9</math>?<br />
<br />
We rearrange the numbers to make <math>(1 + 9) + (2 + 8) + (3 + 7) + (4 + 6) + 5</math>. You might have noticed that each of the terms we put next to each other add up to 10, which makes for easy adding. <math>4 \times 10 + 5 = 45</math>. Adding that on to 900 makes 945.<br />
<br />
945 is <math>\boxed{\text{B}}</math><br />
<br />
===Solution 2===<br />
Instead of breaking the sum and then rearranging, we can start by rearranging:<br />
<cmath>\begin{align*}<br />
90+91+92+\cdots +98+99 &= (90+99)+(91+98)+(92+97)+(93+96)+(94+95) \\<br />
&= 189+189+189+189+189 \\<br />
&= 945\rightarrow \boxed{\text{B}} <br />
\end{align*}</cmath><br />
<br />
===Solution 3===<br />
<br />
We can use a formula. <br />
<br />
It is <math>\frac{n}{2}\times</math> (First term+Last term) where <math>n</math> is the number of terms in the sequence. <br />
<br />
Applying it here:<br />
<br />
<math>\frac{10}{2} \times (90+99) = 945 \rightarrow \boxed{B}</math><br />
<br />
==See Also==<br />
<br />
{{AJHSME box|year=1985|num-b=1|num-a=3}}<br />
<br />
[[Category:Introductory Algebra Problems]]</div>Hrithikguy