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6. 设严格递增的整数数列

a_{1} ， a_{2}, \ldots, a_{20}

满足

a_{1}=1 ， a_{20}=40

. 设

f

为 ...

Mar 8, 2024

6. 设严格递增的整数数列

a_{1} ， a_{2}, \ldots, a_{20}

满足

a_{1}=1 ， a_{20}=40

. 设

f

为

a_{1}+a_{2}, a_{2}+a_{3}, \ldots, a_{19}+a_{20}

这 19 个数中被 3 整除的项的个数，则

f

的最大值为 , 使得

f

取到最大值的数列

\left\{a_{n}\right\}

的个数为

Solution by Steps

step 1

To maximize the number of elements divisible by 3 in the sequence, we need to consider the properties of numbers when they are added

step 2

Since

a_1 = 1

and

a_{20} = 40

, we know that

a_1 + a_2

cannot be divisible by 3 because 1 is not divisible by 3

step 3

To maximize

f

, we want as many pairs

a_i + a_{i+1}

as possible to be divisible by 3. This means we want

a_i

and

a_{i+1}

to be 2 apart, where one is 1 less than a multiple of 3 and the other is 1 more

step 4

We can construct the sequence by choosing

a_2

to be 2 (since

a_1 = 1

), and then ensuring that each subsequent

a_i

is either 1 less or 1 more than a multiple of 3

step 5

The sequence will alternate around multiples of 3, maximizing the number of sums

a_i + a_{i+1}

that are multiples of 3

step 6

To find the maximum value of

f

, we count the number of pairs

a_i + a_{i+1}

that can be made divisible by 3

step 7

Since there are 19 sums and we want to maximize the number divisible by 3, we can have at most 13 sums divisible by 3 because the 14th sum would require

a_{20}

to be 39, which is not allowed as

a_{20} = 40

step 8

To find the number of sequences yielding the maximum

f

, we need to consider the constraints at the beginning and end of the sequence

step 9

There is only one sequence that starts with 1 and ends with 40 while maximizing the number of sums divisible by 3, which is by following the pattern established in steps 4 and 5

Answer

The maximum value of

f

is 13, and the number of sequences

\{a_n\}

yielding the maximum

f

is 1.

Key Concept

Maximizing the number of elements divisible by 3 in a sequence of sums

Explanation

To maximize the number of sums divisible by 3 in a sequence, we arrange the sequence elements around multiples of 3, ensuring that each sum

a_i + a_{i+1}

is as often as possible a multiple of 3, given the constraints at the start and end of the sequence.