一道GRE数学题,OG上的,没明白,Let S be the set of all positive integers n such that n^2 is a multiple of both 24 and108.Which of the following integers are divisors of every integer n in Indicate all such integers.A.12 B.24 C.36 D.72
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![一道GRE数学题,OG上的,没明白,Let S be the set of all positive integers n such that n^2 is a multiple of both 24 and108.Which of the following integers are divisors of every integer n in Indicate all such integers.A.12 B.24 C.36 D.72](/uploads/image/z/8981436-12-6.jpg?t=%E4%B8%80%E9%81%93GRE%E6%95%B0%E5%AD%A6%E9%A2%98%2COG%E4%B8%8A%E7%9A%84%2C%E6%B2%A1%E6%98%8E%E7%99%BD%2CLet+S+be+the+set+of+all+positive+integers+n+such+that+n%5E2+is+a+multiple+of+both+24+and108.Which+of+the+following+integers+are+divisors+of+every+integer+n+in+Indicate+all+such+integers.A.12+B.24+C.36+D.72)
一道GRE数学题,OG上的,没明白,Let S be the set of all positive integers n such that n^2 is a multiple of both 24 and108.Which of the following integers are divisors of every integer n in Indicate all such integers.A.12 B.24 C.36 D.72
一道GRE数学题,OG上的,没明白,
Let S be the set of all positive integers n such that n^2 is a multiple of both 24 and108.Which of the following integers are divisors of every integer n in Indicate all such integers.
A.12 B.24 C.36 D.72
一道GRE数学题,OG上的,没明白,Let S be the set of all positive integers n such that n^2 is a multiple of both 24 and108.Which of the following integers are divisors of every integer n in Indicate all such integers.A.12 B.24 C.36 D.72
先理解题目:S是所有正整数n的集合,n的特点是n^2是24和108的公倍数.问选项里哪些数是S里所有n的公约数.
24和108的公倍数是12×2×9×a,a是一个正整数.
所以n^2=12×2×9×a,n = sqrt(12×2×9×a) = 6×sqrt(6×a),a应该等于6,6×4,6×9,6×16,6×25,6×m^2(m是整数)之类的,这样才能保证sqrt(6×a)能开根号得到整数.所以n=36,72,108,36×m.n最小是36.
所以答案是AC.
意思是S是满足n^2是24和108的公倍数的所有正整数,下面选项哪一个是n的除数。
解题步骤:由题知n平方是24和108的倍数,24=2*2*2*3,108=3*3*3*2*2,n平方最小为3*3*3*2*2*2=216,开方后得6*根号6,而集合是正整数集合,所以集合中最小的数为6*根号6*根号6=36,其他数应是36的倍数,由此可得集合中的公约数应有12和36...
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意思是S是满足n^2是24和108的公倍数的所有正整数,下面选项哪一个是n的除数。
解题步骤:由题知n平方是24和108的倍数,24=2*2*2*3,108=3*3*3*2*2,n平方最小为3*3*3*2*2*2=216,开方后得6*根号6,而集合是正整数集合,所以集合中最小的数为6*根号6*根号6=36,其他数应是36的倍数,由此可得集合中的公约数应有12和36
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