AMC test question

  • Thread starter cdhotfire
  • Start date
  • #1
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This is not exatly homework, but it seems the best place to post it.

Let A, M, and C be digits with:

(100A+10M+C)(A+M+C) = 2005

A. 1 B. 2 C. 3 D. 4 E. 5

maybe some one can help me out on this, was one of the test questions i did not know how to do. :blushing:
 

Answers and Replies

  • #2
dextercioby
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I don't see other solution than
[tex] A=4;M=0;C=1 [/tex]

Daniel.
 
  • #3
Use the fact that 2005 is written in decimal notation, which is incidently convenient with respect to how the left-hand side is written. Remember how decimal notation is defined. Any ideas ?
 
  • #4
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wow, i didnt notice that, hmmm, thxs. :smile:
 
  • #5
t!m
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Yeah dex is right. If you look at the range of answers, assuming A, M, and C are all going to be in that range, then 2005 only divides evenly by 5, leaving 401. Quick glance yields 4, 0, and 1.
 
  • #6
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how about this one:

Three circles of radius [itex]s[/itex] are drawn in the first quadrant of the xy-plane. The first circle is tangent to both axes, the second is tangent to the first circle and the x-axis, and the third is tangent to the first circle and the y-axis. A circle of radius [itex] r > s [/itex] is tangent to both axes and to the second and third circles. What is [itex] r/s [/itex]?

A. 5 B. 6 C.8 D. 9 E. 10
 
  • #7
t!m said:
Yeah dex is right. If you look at the range of answers, assuming A, M, and C are all going to be in that range, then 2005 only divides evenly by 5, leaving 401. Quick glance yields 4, 0, and 1.
That is a better method. :smile:
 
  • #8
t!m
147
6
cdhotfire: Here's how I did it.

I drew a line from the center of the large circle to the center of the top small circle (tangent to the first circle and the y-axis) and used that as the hypotenuse for a right triangle with sides (r-3s), (r-s), and (r+s). This makes more sense if you're looking at the diagram. In which case, (r-s)^2 + (r-3s)^2 = (r+s)^2. I just tried each answer, and found r=9, s=1 (r/s=9) to be the one which works. If you're confused, I'll see if I can work up a diagram.
 
  • #9
193
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thxs man, that really helped me out. :). How about the star one:

In the five-sided star shown, the letters A, B, C, D and E are replaced by the numbers 3, 5, 6, 7 and 9, although not necessarily in that order. The sums of the numbers at the ends of the line segments [itex]\lin{AB},\lin{BC},\lin{CD},\lin{DE}[/itex] and [itex]\lin{EA}[/itex] form an arithmetic sequence, although not necessarily in that order. What is the middle term of the arithmetic sequence?

http://img103.exs.cx/img103/6466/amc7lx.gif

A. 9 B. 10 C. 11 D. 12 E. 13
 
  • #10
t!m
147
6
Put down the nine arbitrarily. Connect it to the two smallest, 3 and 5. Then to make the 3 connected to the two largest, connect it to 7 in addition to 9. Put the 5 in the remaining spot and you have 10, 11, 12, 13, and 14, with 12 being the middle term.
 
  • #11
193
0
o i see, i didnt understand the question. :rolleyes: , thxs though. :smile:
 

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