Difference between revisions of "2010 AIME II Problems/Problem 12"

Revision as of 05:10, 13 October 2023

Problem

For how many different digits $n$ is the three-digit number $14n$ divisible by $n$ ?

Note: $14n$ refers to a three-digit number with the unit digit of $n,$ not the product of $14$ and $n.$

Solution 1

For how many different digits $n$ is the three-digit number $14n$ divisible by $n$ ?

Note: $14n$ refers to a three-digit number with the unit digit of $n,$ not the product of $14$ and $n.$

Solution 2

Let $s$ be the semiperimeter of the two triangles. Also, let the base of the longer triangle be $16x$ and the base of the shorter triangle be $14x$ for some arbitrary factor $x$ . Then, the dimensions of the two triangles must be $s-8x,s-8x,16x$ and $s-7x,s-7x,14x$ . By Heron's Formula, we have

$\sqrt{s(8x)(8x)(s-16x)}=\sqrt{s(7x)(7x)(s-14x)}$ $8\sqrt{s-16x}=7\sqrt{s-14x}$ $64s-1024x=49s-686x$ $15s=338x$

Since $15$ and $338$ are coprime, to minimize, we must have $s=338$ and $x=15$ . However, we want the minimum perimeter. This means that we must multiply our minimum semiperimeter by $2$ , which gives us a final answer of $\boxed{676}$ .

Solution 3

Let the first triangle have sides $16n,a,a$ , so the second has sides $14n,a+n,a+n$ . The height of the first triangle is $\frac{7}{8}$ the height of the second triangle. Therefore, we have $a^2-64n^2=\frac{49}{64}((a+n)^2-49n^2).$ Multiplying this, we get $64a^2-4096n^2=49a^2+98an-2352n^2,$ which simplifies to $15a^2-98an-1744n^2=0.$ Solving this for $a$ , we get $a=n\cdot\frac{218}{15}$ , so $n=15$ and $a=218$ and the perimeter is $15\cdot16+218+218=\boxed{676}$ .

~john0512

Note

We use $16x$ and $14x$ instead of $8x$ and $7x$ to ensure that the triangle has integral side lengths. Plugging $8x$ and $7x$ directly into Heron's gives $s=338$ , but for this to be true, the second triangle would have side lengths of $\frac{223}{2}$ , which is impossible.

~jd9

@@ Line 6: / Line 6: @@
 == Solution 1==
-Let the first triangle have side lengths <math>a</math>, <math>a</math>, <math>14c</math>, and the second triangle have side lengths <math>b</math>, <math>b</math>, <math>16c</math>, where <math>a, b, 2c \in \mathbb{Z}</math>.
+For how many different digits <math>n</math> is the three-digit number <math>14n</math> divisible by <math>n</math>?
-<br/>
+Note: <math>14n</math> refers to a three-digit number with the unit digit of <math>n,</math> not the product of <math>14</math> and <math>n.</math>
-Equal perimeter:
-<center>
-<math>\begin{array}{ccc}
-a+14c&=&2b+16c\\
-a+7c&=&b+8c\\
-c&=&a-b\\
-\end{array}</math>
-</center>
-<br/>
-Equal Area:
-<center>
-<math>\begin{array}{cccl}
-c(\sqrt{a^2-(7c)^2})&=&8c(\sqrt{b^2-(8c)^2})&{}\\
-(\sqrt{(a+7c)(a-7c)})&=&8(\sqrt{(b+8c)(b-8c)})&{}\\
-(\sqrt{(a-7c)})&=&8(\sqrt{(b-8c)})&\text{(Note that } a+7c=b+8c)\\
-a-343c&=&64b-512c&{}\\
-a+169c&=&64b&{}\\
-a+169(a-b)&=&64b&\text{(Note that } c=a-b)\\
-a&=&233b&{}\\
-\end{array}</math>
-</center>
-Since <math>a</math> and <math>b</math> are integer, the minimum occurs when <math>a=233</math>, <math>b=218</math>, and <math>c=15</math>. Hence, the perimeter is <math>2a+14c=2(233)+14(15)=\boxed{676}</math>.
 == Solution 2==

2010 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 11	Followed by Problem 13
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