Difference between revisions of "2008 AIME I Problems/Problem 3"

Revision as of 20:52, 24 April 2021

Problem

Ed and Sue bike at equal and constant rates. Similarly, they jog at equal and constant rates, and they swim at equal and constant rates. Ed covers $74$ kilometers after biking for $2$ hours, jogging for $3$ hours, and swimming for $4$ hours, while Sue covers $91$ kilometers after jogging for $2$ hours, swimming for $3$ hours, and biking for $4$ hours. Their biking, jogging, and swimming rates are all whole numbers of kilometers per hour. Find the sum of the squares of Ed's biking, jogging, and swimming rates.

Solution

Solution 1

Let the biking rate be $b$ , swimming rate be $s$ , jogging rate be $j$ , all in km/h.

We have $2b + 3j + 4s = 74,2j + 3s + 4b = 91$ . Subtracting the second from twice the first gives $4j + 5s = 57$ . Mod 4, we need $s\equiv1\pmod{4}$ . Thus, $(j,s) = (13,1),(8,5),(3,9)$ .

$(13,1)$ and $(3,9)$ give non-integral $b$ , but $(8,5)$ gives $b = 15$ . Thus, our answer is $15^{2} + 8^{2} + 5^{2} = \boxed{314}$ .

Solution 2

Let $b$ , $j$ , and $s$ be the biking, jogging, and swimming rates of the two people. Hence, $2b + 3j + 4s = 74$ and $4b + 2j + 3s = 91$ . Subtracting gives us that $2b - j - s = 17$ . Adding three times this to the first equation gives that $8b + s = 125\implies b\le 15$ . Adding four times the previous equation to the first given one gives us that $10b - j = 142\implies b > 14\implies b\ge 15$ . This gives us that $b = 15$ , and then $j = 8$ and $s = 5$ . Therefore, $b^2 + s^2 + j^2 = 225 + 64 + 25 = \boxed{314}$ .

Solution 3

Creating two systems, we get $2x+3y+4z=74$ , and $2y+3z+4x=91$ . Subtracting the two expressions we get $y+z-2x=-17$ . Note that $-17$ is odd, so one of $x,y,z$ is odd. We see from our second expression that $z$ must be odd, because $91$ is also odd and $2y$ and $4x$ are odd. Thus, with this information, we can test cases quickly:

When readdressing the first equation, we see that if $2x+3y$ will be a multiple of $6$ , $4z \equiv 2 \pmod{6} = 5$ , we get that $x=15$ and $y=8$ , which works because of integer values. Therefore, $225+64+25=\boxed{314}$

@@ Line 13: / Line 13: @@
 === Solution 2 ===
 Let <math>b</math>, <math>j</math>, and <math>s</math> be the biking, jogging, and swimming rates of the two people.  Hence, <math>2b + 3j + 4s = 74</math> and <math>4b + 2j + 3s = 91</math>.  Subtracting gives us that <math>2b - j - s = 17</math>.  Adding three times this to the first equation gives that <math>8b + s = 125\implies b\le 15</math>.  Adding four times the previous equation to the first given one gives us that <math>10b - j = 142\implies b > 14\implies b\ge 15</math>.  This gives us that <math>b = 15</math>, and then <math>j = 8</math> and <math>s = 5</math>.  Therefore, <math>b^2 + s^2 + j^2 = 225 + 64 + 25 = \boxed{314}</math>.
+=== Solution 3 ===
+Creating two systems, we get <math>2x+3y+4z=74</math>, and <math>2y+3z+4x=91</math>. Subtracting the two expressions we get <math>y+z-2x=-17</math>. Note that <math>-17</math> is odd, so one of <math>x,y,z</math> is odd. We see from our second expression that <math>z</math> must be odd, because <math>91</math> is also odd and <math>2y</math> and <math>4x</math> are odd. Thus, with this information, we can test cases quickly:
+When readdressing the first equation, we see that if <math>2x+3y</math> will be a multiple of <math>6</math>, <math>4z \equiv 2 \pmod{6} = 5</math>, we get that <math>x=15</math> and <math>y=8</math>, which works because of integer values. Therefore, <math>225+64+25=\boxed{314}</math>
 == See also ==

2008 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 2	Followed by Problem 4
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