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Difference between revisions of "2024 DMC Mock 10 Problems/Problem 8"

(Created page with "Note the identity <math>a^2+b^2+c^2-ab-ba-ca=\frac{(a-b)^2+(b-c)^2+(c-a)^2}{2}</math>. Substituting in the values, we see that the expression is <math>\frac{3^2+4^2+7^2}{2}=\b...")
 
 
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Note the identity <math>a^2+b^2+c^2-ab-ba-ca=\frac{(a-b)^2+(b-c)^2+(c-a)^2}{2}</math>. Substituting in the values, we see that the expression is <math>\frac{3^2+4^2+7^2}{2}=\boxed{37}</math>. Alternatively, notice that the expression is 2\pmod{5}, so the answer must be <math>\boxed{B}</math>.
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Note the identity <math>a^2+b^2+c^2-ab-ba-ca=\frac{(a-b)^2+(b-c)^2+(c-a)^2}{2}</math>. Substituting in the values, we see that the expression is <math>\frac{3^2+4^2+7^2}{2}=\boxed{37}</math>. Alternatively, notice that the expression is 2<math>\pmod{5}</math>, so the answer must be <math>\boxed{B}</math>.

Latest revision as of 20:03, 16 September 2024

Note the identity $a^2+b^2+c^2-ab-ba-ca=\frac{(a-b)^2+(b-c)^2+(c-a)^2}{2}$. Substituting in the values, we see that the expression is $\frac{3^2+4^2+7^2}{2}=\boxed{37}$. Alternatively, notice that the expression is 2$\pmod{5}$, so the answer must be $\boxed{B}$.

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