Difference between revisions of "2006 AMC 12A Problems/Problem 9"
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== Problem == | == Problem == | ||
− | Oscar buys <math>13</math> pencils and <math>3</math> erasers for <math> | + | Oscar buys <math>13</math> pencils and <math>3</math> erasers for <math></math>1.00<math>. A pencil costs more than an eraser, and both items cost a [[whole number]] of cents. What is the total cost, in cents, of one pencil and one eraser? |
</math> \mathrm{(A) \ } 10\qquad \mathrm{(B) \ } 12\qquad \mathrm{(C) \ } 15\qquad \mathrm{(D) \ } 18<math> | </math> \mathrm{(A) \ } 10\qquad \mathrm{(B) \ } 12\qquad \mathrm{(C) \ } 15\qquad \mathrm{(D) \ } 18<math> |
Revision as of 19:23, 5 January 2008
Problem
Oscar buys pencils and erasers for $$ (Error compiling LaTeX. )1.00 \mathrm{(A) \ } 10\qquad \mathrm{(B) \ } 12\qquad \mathrm{(C) \ } 15\qquad \mathrm{(D) \ } 18$$ (Error compiling LaTeX. )\mathrm{(E) \ } 20pe13p + 3e = 100p > e > 0pee \geq 1p \geq 2$.
Considering the [[equation]]$ (Error compiling LaTeX. )13p + 3e = 100p + 0e \equiv 1 \pmod 3p$leaves a remainder of 1 on division by 3.
Since$ (Error compiling LaTeX. )p \geq 2p$are 4, 7, 10 ....
Since 13 pencils cost less than 100 cents,$ (Error compiling LaTeX. )13p < 10013 \times 10 = 130p$must be 4 or 7.
If$ (Error compiling LaTeX. )p = 413p = 523e = 48e = 16p7 \times 13= 91\frac{9}{3} = 3$cents.
Thus one pencil plus one eraser cost$ (Error compiling LaTeX. )7 + 3 = 10\mathrm{(A) \ }$.
See also
2006 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 8 |
Followed by Problem 10 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |