A set S consists of n consecutive positive integers, where n > 1 is an odd integer. If the… — GMAT

A set $S$ consists of $n$ consecutive positive integers, where $n > 1$ is an odd integer. If the sum of all elements in set $S$ is equal to $3^{10}$ , what is the minimum possible value of the median of set $S$ ?

Cevap: 243

Cevap

The minimum possible value of the median of set

S

is 243.

For an odd number

n

of consecutive integers, the sum of the set equals

n \times m

, where

m

is the median. Given that

n \times m = 3^{10}

, both

n

and

m

must be powers of 3, so

n = 3^k

and

m = 3^{10-k}

. To ensure all terms in the set are positive, the smallest term

m - \frac{n-1}{2}

must be at least 1, which requires

2m > n

. Substituting the powers of 3 yields

2 \cdot 3^{10-k} > 3^k \implies 3^{2k} < 2 \cdot 3^{10} = 118,098

. The largest integer

k

satisfying this condition is

k = 5

(since

3^{10} = 59,049 < 118,098

while

3^{12} = 531,441 > 118,098

). Maximizing

k

minimizes the median

m = 3^{10-5} = 3^5 = 243

Adım Adım Çözüm

Relate the sum of an evenly spaced set to its number of terms and median.

n \times m = 3^{10}

, where

n

is the number of terms and

m

is the median.

For any set of

n

consecutive integers where

n

is odd, the sum of the set equals the number of terms times the middle term (median).

Express

n

and

m

as powers of 3.

n = 3^k

and

m = 3^{10-k}

where

k \ge 1

Because

3^{10}

has only 3 as a prime factor, any integer factors

n

and

m

must be powers of 3.

Establish the positivity constraint for the terms in set

S

2m > n

The smallest term in the set is

m - \frac{n-1}{2}

. Requiring

m - \frac{n-1}{2} \ge 1

gives

2m \ge n + 1

, or strictly

2m > n

Solve the inequality

2m > n

in terms of

k

3^{2k} < 2 \cdot 3^{10} = 118,098

Substituting

n = 3^k

and

m = 3^{10-k}

yields

2 \cdot 3^{10-k} > 3^k

, which rearranges to

3^{2k} < 2 \cdot 3^{10}

Find the maximum valid integer value of

k

and calculate the corresponding minimum median

m

Maximum

k = 5

, giving minimum median

m = 3^{5} = 243

For

k = 5

3^{10} = 59,049 < 118,098

. For

k = 6

3^{12} = 531,441 > 118,098

. Thus

k=5

is the maximum integer

k

, which yields the minimum median

m = 3^{10-5} = 243

Anahtar Kavram

Sum of consecutive integers set formula and positivity constraints

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