HMMT 二月 2006 · 冲刺赛 · 第 27 题
HMMT February 2006 — Guts Round — Problem 27
题目详情
- [9] Let N denote the number of subsets of { 1 , 2 , 3 , . . . , 100 } that contain more prime numberskthan multiples of 4. Compute the largest integer k such that 2 divides N .
IX HARVARD-MIT MATHEMATICS TOURNAMENT, 25 FEBRUARY 2006 — GUTS ROUNDth
英文原题
[9] Let N denote the number of subsets of { 1 , 2 , 3 , . . . , 100 } that contain more prime numbers
than multiples of 4. Compute the largest integer k such that 2 k divides N .
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IX th HARVARD-MIT MATHEMATICS TOURNAMENT, 25 FEBRUARY 2006 — GUTS ROUND
解析
英文解析
- Let N denote the number of subsets of { 1 , 2 , 3 , . . . , 100 } that contain more primeknumbers than multiples of 4. Compute the largest integer k such that 2 divides N .
Answer: 52
Solution: Let S denote a subset with the said property. Note that there are 25
multiples of 4 and 25 primes in the set { 1 , 2 , 3 , . . . , 100 } , with no overlap between thetwo. Let T denote the subset of 50 numbers that are neither prime nor a multiple of 4,
and let U denote the 50 other numbers. Elements of T can be arbitrarily included inor excluded by S . Consider S ∩ U = S and U − S = S (the set difference is defined
1 2
to be all elements of U that are not in S .) S and S are two disjoint sets such that
1 2
U = S ∩ S . If S contains more multiples of 4 than primes, then S contains more
1 2 1 2
primes than multiples of 4, and conversely. Furthermore, S contains an equal numberof primes and multiples of 4 if and only if S contains equal numbers as well. Let V1
denote an arbitrary subset of T . It follows from examining pairs of sets V ∪ S and 2
V ∪ S that 1
( )2
( )
∑225
1 25
50 50
N = 2 · 2 −
2 kk =0
( ( ))
49 5050 = 2 · 2 −
Since 50! is divisible by 2 exactly 25 + 12 + 6 + 3 + 1 = 47 times while 25! is divisible 25
( )
by 2 exactly 12 + 6 + 3 + 1 = 22 times, it follows that is divisible by 2 exactly 350
times, so the answer is 49 + 3 = 52.25