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AMC8，--，2006年真题解析( 四 )


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Answer： C
Solution：
Since there are 6 players ，a total of 6*5/2=15 games are played. So far ，4+3+2+2+2=13 games finished （one person won from each game） so Monica needs to win 15-13=2.


AMC8，--，2006年真题解析
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Answer： A
Solution：
The water level will rise 1cm for every 100*40=4 ， 000 . Since 1 ， 000 is 1/4 of 4 ， 000 ，the water will rise 1/4=0.25.


AMC8，--，2006年真题解析
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AMC8，--，2006年真题解析
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Answer： D
Solution：
If the lower cells contain A ， B ，and C ，then the second row will contain A+B and B+C，and the top cell will contain A+2B+C. To obtain the smallest sum ，place 1 in the center cell and 2 and 3 in the outer ones. The top number will be 7. For the largest sum ，place 9 in the center cell and 7 and 8 in the outer ones. This top number will be 33. The difference is 33-7=26.


AMC8，--，2006年真题解析
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Answer： A
Solution：
x=6*q1+4； => x+2=6*（q1+1）
x=5*q2+3； =>x+2=5*（q2+1）
If there were two more coins in the box ，the number of coins would be divisible by both 6 and 5. The smallest number that is divisible by 6 and 5 is 30 ，so the smallest possible number of coins in the box is 28 and the remainder when divided by 7 is 0.


AMC8，--，2006年真题解析
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Answer： A
Solution：
CD*101=CDCD. So ABA=101 ，A=1 ， b=0. A+B=1.


AMC8，--，2006年真题解析
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Answer： B
Solution：
Notice that 44 and 38 are both even ，while 59 is odd. If any odd prime is added to 59 ，an even number will be obtained. However ，the only way to obtain this even number would be to add another even number to 44 ，and a different one to 38. Since there is only one even prime （2） the middle card“s hidden number cannot be an odd prime ，and so must be even. Therefore ，the middle card”s hidden number must be 2 ，so the constant sum is 59+2=61. Thus ，the first card“s hidden number is 61-44=17 ，and the last card”s hidden number is 61-38=23. Since the sum of the hidden primes is 2+17+23=42. the average of the primes is 42/3=14.


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