2018-08-31 重点归纳
AMC10数学竞赛是美国高中数学竞赛中的一项,是针对高中一年级及初中三年级学生的数学测试,该竞赛开始于2000年,分A赛和B赛,于每年的2月初和2月中举行,学生可任选参加一项即可。不管是对高校申请还是今后在数学领域的发展都极其有利!那么接下来跟随小编来看一下AMC10数学竞赛真题以及官方解答吧:
Last year Isabella took  math tests and received
 math tests and received  different scores, each an integer between
 different scores, each an integer between  and
 and  , inclusive. After each test she noticed that the average of her test scores was an integer. Her score on the seventh test was
, inclusive. After each test she noticed that the average of her test scores was an integer. Her score on the seventh test was  . What was her score on the sixth test?
. What was her score on the sixth test?

Let the sum of the scores of Isabella's first  tests be
 tests be  . Since the mean of her first
. Since the mean of her first  scores is an integer, then
 scores is an integer, then  , or
, or . Also,
. Also,  , so by CRT,
, so by CRT,  . We also know that
. We also know that  , so by inspection,
, so by inspection,  . However, we also have that the mean of the first
. However, we also have that the mean of the first  integers must be an integer, so the sum of the first
 integers must be an integer, so the sum of the first  test scores must be an multiple of
 test scores must be an multiple of  , which implies that the
, which implies that the  th test score is
th test score is  .
.
First, we find the largest sum of scores which is  which equals
 which equals  . Then we find the smallest sum of scores which is
. Then we find the smallest sum of scores which is  which is
 which is  . So the possible sums for the 7 test scores so that they provide an integer average are
. So the possible sums for the 7 test scores so that they provide an integer average are  and
 and  which are
 which are  and
 and  respectively. Now in order to get the sum of the first 6 tests, we negate
 respectively. Now in order to get the sum of the first 6 tests, we negate  from each sum producing
 from each sum producing  and
 and  . Notice only
. Notice only  is divisible by
 is divisible by  so, therefore, the sum of the first
 so, therefore, the sum of the first  tests is
 tests is  . We need to find her score on the
. We need to find her score on the  test so what number minus
 test so what number minus  will give us a number divisible by
 will give us a number divisible by  . Since
. Since  is the
 is the  test score and all test scores are distinct that only leaves
 test score and all test scores are distinct that only leaves  .
.
By inspection, the sequences  and
 and  work, so the answer is
 work, so the answer is  . Note: A method of finding this "cheap" solution is to create a "mod chart", basically list out the residues of 91-100 modulo 1-7 and then finding the two sequences should be made substantially easier.
. Note: A method of finding this "cheap" solution is to create a "mod chart", basically list out the residues of 91-100 modulo 1-7 and then finding the two sequences should be made substantially easier.
Since all of the scores are from  , we can 'subtract' 90 off from all of the scores. Basically, we're looking at the units digits except for 100; we're looking at 10 in this case. Since the last score was a 95, the sum of the scores from the first six tests must be
, we can 'subtract' 90 off from all of the scores. Basically, we're looking at the units digits except for 100; we're looking at 10 in this case. Since the last score was a 95, the sum of the scores from the first six tests must be  and
 and  . Trying out a few cases, the only solution possible is 30 (this is from adding numbers 1-10). The sixth test score must be
. Trying out a few cases, the only solution possible is 30 (this is from adding numbers 1-10). The sixth test score must be  because
 because  . The only possible test scores are
. The only possible test scores are  and
 and  , so the answer is
, so the answer is  .
.
以上就是小编对AMC10数学竞赛真题以及解析的介绍,希望对你有所帮助,更多学习资料请持续关注AMC数学竞赛网!
 
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