https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Hrithikguy&feedformat=atom AoPS Wiki - User contributions [en] 2021-04-22T18:17:49Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2011_USAJMO_Problems/Problem_1&diff=38263 2011 USAJMO Problems/Problem 1 2011-04-29T00:43:53Z <p>Hrithikguy: </p> <hr /> <div>Find, with proof, all positive integers &lt;math&gt;n&lt;/math&gt; for which &lt;math&gt;2^n + 12^n + 2011^n&lt;/math&gt; is a perfect square.<br /> <br /> <br /> <br /> Let &lt;math&gt;2^n + 12^n + 2011^n = x^2&lt;/math&gt;<br /> &lt;math&gt;(-1)^n + 1 \equiv x^2 \pmod {3}&lt;/math&gt;.<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 &lt;math&gt;n&lt;/math&gt; that satisfies is &lt;math&gt;n = 1&lt;/math&gt;.<br /> Assume that &lt;math&gt;n \ge 2&lt;/math&gt;. <br /> Then consider the equation<br /> &lt;math&gt;2^n + 12^n = x^2 - 2011^n&lt;/math&gt;.<br /> From modulo 2, we easily that x is odd. Let &lt;math&gt;x = 2a + 1&lt;/math&gt;, where a is an integer.<br /> &lt;math&gt;2^n + 12^n = 4a^2 + 4a + 1 - 2011^n&lt;/math&gt;.<br /> Dividing by 4,<br /> &lt;math&gt;2^{n-2} + 3^n \cdot 4^{n-1} = a^2 + a + \dfrac {1}{4} (1 - 2011^n})&lt;/math&gt;.<br /> Since &lt;math&gt;n \ge 2&lt;/math&gt;, &lt;math&gt;n-2 \ge 0&lt;/math&gt;, so &lt;math&gt;2^{n-2}&lt;/math&gt; similarly, the entire LHS is an integer, and so are &lt;math&gt;a^2&lt;/math&gt; and &lt;math&gt;a&lt;/math&gt;. Thus, &lt;math&gt; \dfrac {1}{4} (1 - 2011^n})&lt;/math&gt; must be an integer.<br /> Let &lt;math&gt; \dfrac {1}{4} (1 - 2011^n}) = k&lt;/math&gt;. Then we have &lt;math&gt;1- 2011^n = 4k&lt;/math&gt;.<br /> &lt;math&gt;1- 2011^n \equiv 0 \pmod {4}&lt;/math&gt;<br /> &lt;math&gt;(-1)^n \equiv 1 \pmod {4}&lt;/math&gt;.<br /> Thus, n is even.<br /> However, I have already shown that &lt;math&gt;n&lt;/math&gt; must be odd. This is a contradiction. Therefore, &lt;math&gt;n&lt;/math&gt; 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> Hrithikguy https://artofproblemsolving.com/wiki/index.php?title=2011_USAJMO_Problems/Problem_1&diff=38262 2011 USAJMO Problems/Problem 1 2011-04-29T00:39:19Z <p>Hrithikguy: Created page with 'Let &lt;math&gt;2^n + 12^n + 2011^n = x^2&lt;/math&gt; &lt;math&gt;(-1)^n + 1 \equiv x^2 \pmod {3}&lt;/math&gt;. Since all perfect squares are congruent to 0 or 1 modulo 3, this means that n must be odd…'</p> <hr /> <div>Let &lt;math&gt;2^n + 12^n + 2011^n = x^2&lt;/math&gt;<br /> &lt;math&gt;(-1)^n + 1 \equiv x^2 \pmod {3}&lt;/math&gt;.<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 &lt;math&gt;n&lt;/math&gt; that satisfies is &lt;math&gt;n = 1&lt;/math&gt;.<br /> Assume that &lt;math&gt;n \ge 2&lt;/math&gt;. <br /> Then consider the equation<br /> &lt;math&gt;2^n + 12^n = x^2 - 2011^n&lt;/math&gt;.<br /> From modulo 2, we easily that x is odd. Let &lt;math&gt;x = 2a + 1&lt;/math&gt;, where a is an integer.<br /> &lt;math&gt;2^n + 12^n = 4a^2 + 4a + 1 - 2011^n&lt;/math&gt;.<br /> Dividing by 4,<br /> &lt;math&gt;2^{n-2} + 3^n \cdot 4^{n-1} = a^2 + a + \dfrac {1}{4} (1 - 2011^n})&lt;/math&gt;.<br /> Since &lt;math&gt;n \ge 2&lt;/math&gt;, &lt;math&gt;n-2 \ge 0&lt;/math&gt;, so &lt;math&gt;2^{n-2}&lt;/math&gt; similarly, the entire LHS is an integer, and so are &lt;math&gt;a^2&lt;/math&gt; and &lt;math&gt;a&lt;/math&gt;. Thus, &lt;math&gt; \dfrac {1}{4} (1 - 2011^n})&lt;/math&gt; must be an integer.<br /> Let &lt;math&gt; \dfrac {1}{4} (1 - 2011^n}) = k&lt;/math&gt;. Then we have &lt;math&gt;1- 2011^n = 4k&lt;/math&gt;.<br /> &lt;math&gt;1- 2011^n \equiv 0 \pmod {4}&lt;/math&gt;<br /> &lt;math&gt;(-1)^n \equiv 1 \pmod {4}&lt;/math&gt;.<br /> Thus, n is even.<br /> However, I have already shown that &lt;math&gt;n&lt;/math&gt; must be odd. This is a contradiction. Therefore, &lt;math&gt;n&lt;/math&gt; 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> Hrithikguy https://artofproblemsolving.com/wiki/index.php?title=1985_AJHSME_Problems/Problem_2&diff=34592 1985 AJHSME Problems/Problem 2 2010-06-05T02:42:30Z <p>Hrithikguy: /* Solution 3 */</p> <hr /> <div>==Problem==<br /> <br /> &lt;math&gt;90+91+92+93+94+95+96+97+98+99=&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;\text{(A)}\ 845 \qquad \text{(B)}\ 945 \qquad \text{(C)}\ 1005 \qquad \text{(D)}\ 1025 \qquad \text{(E)}\ 1045&lt;/math&gt;<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 -&gt; &lt;math&gt;90 + 91 + ... + 98 + 99 = 90 \times 10 + 1 + 2 + 3 + ... + 8 + 9&lt;/math&gt;<br /> <br /> We know &lt;math&gt;90 \times 10&lt;/math&gt;, that's easy - &lt;math&gt;900&lt;/math&gt;. So how do we find &lt;math&gt;1 + 2 + ... + 8 + 9&lt;/math&gt;?<br /> <br /> We rearrange the numbers to make &lt;math&gt;(1 + 9) + (2 + 8) + (3 + 7) + (4 + 6) + 5&lt;/math&gt;. You might have noticed that each of the terms we put next to each other add up to 10, which makes for easy adding. &lt;math&gt;4 \times 10 + 5 = 45&lt;/math&gt;. Adding that on to 900 makes 945.<br /> <br /> 945 is &lt;math&gt;\boxed{\text{B}}&lt;/math&gt;<br /> <br /> ===Solution 2===<br /> Instead of breaking the sum and then rearranging, we can start by rearranging:<br /> &lt;cmath&gt;\begin{align*}<br /> 90+91+92+\cdots +98+99 &amp;= (90+99)+(91+98)+(92+97)+(93+96)+(94+95) \\<br /> &amp;= 189+189+189+189+189 \\<br /> &amp;= 945\rightarrow \boxed{\text{B}} <br /> \end{align*}&lt;/cmath&gt;<br /> <br /> ===Solution 3===<br /> <br /> We can use a formula. <br /> <br /> It is &lt;math&gt;\frac{n}{2}\times&lt;/math&gt; (First term+Last term) where &lt;math&gt;n&lt;/math&gt; is the number of terms in the sequence. <br /> <br /> Applying it here:<br /> <br /> &lt;math&gt;\frac{10}{2} \times (90+99) = 945 \rightarrow \boxed{B}&lt;/math&gt;<br /> <br /> ==See Also==<br /> <br /> {{AJHSME box|year=1985|num-b=1|num-a=3}}<br /> <br /> [[Category:Introductory Algebra Problems]]</div> Hrithikguy