2003 AMC 12A Problems/Problem 5: Difference between revisions

Revision as of 10:13, 11 November 2006

The sume of the two 5-digit numbers $AMC10$ and $AMC12$ is $123422$ . What is $A+M+C$ ?

$\mathrm{(A) \ } 10\qquad \mathrm{(B) \ } 11\qquad \mathrm{(C) \ } 12\qquad \mathrm{(D) \ } 13\qquad \mathrm{(E) \ } 14$

$AMC10+AMC12=123422$

$AMC00+AMC00=123400$

$AMC+AMC=1234$

$2\cdot AMC=1234$

$AMC=\frac{1234}{2}=617$

Since $A$ , $M$ , and $C$ are digits, $A=6$ , $M=1$ , $C=7$ .

Therefore, $A+M+C = 6+1+7 = 14 \Rightarrow E$ .

Revision as of 22:42, 9 November 2006 view source Xantos C. Guin (talk \| contribs) editor 580 edits added problem and solution	Revision as of 10:13, 11 November 2006 view source JBL (talk \| contribs) Administrators 2,310 edits m 2003 AMC 12A/Problem 5 moved to 2003 AMC 12A Problems/Problem 5 Newer edit →
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