Solved on Feb 14, 2024

Find the integer between 9 and 999 with the most positive divisors.

STEP 1

Assumptions
1. We are looking for a number between 9 and 999.
2. We want to find the number with the most factors within this range.
3. A factor of a number is an integer that can divide the number without leaving a remainder.
4. The number with the most factors is not necessarily a prime number, as prime numbers have only two factors.

STEP 2

Understand that the number of factors a number has is related to its prime factorization. The prime factorization of a number is the expression of the number as a product of its prime factors.

STEP 3

Recall that the number of factors of a number can be determined by adding 1 to each of the exponents in its prime factorization and then multiplying these numbers together.
$\text{If } n = p_1^{a_1} \cdot p_2^{a_2} \cdot \ldots \cdot p_k^{a_k}, \text{ then the number of factors of } n \text{ is } (a_1 + 1)(a_2 + 1) \ldots (a_k + 1)$
where $p_1, p_2, \ldots, p_k$ are the prime factors of $n$ and $a_1, a_2, \ldots, a_k$ are their respective exponents.

STEP 4

Note that to maximize the number of factors, we want to maximize the product of the exponents (plus one). Since the number of factors increases with the number of distinct prime factors and their powers, we should look for numbers that have a prime factorization with small primes raised to higher powers.

STEP 5

Start by considering the highest power of 2 that is less than 999, since 2 is the smallest prime number. This is $2^9 = 512$ .

STEP 6

Check the number of factors for $2^9$ . Since it has only one prime factor, the number of factors is $9 + 1 = 10$ .

STEP 7

Now consider the product of the smallest primes, such as $2 \times 3 = 6$ , and find the highest power of this product that is less than 999. This is $6^4 = 1296$ , which is greater than 999, so we need to consider $6^3 = 216$ .

STEP 8

Check the number of factors for $6^3$ , which is $2^3 \times 3^3$ . The number of factors is $(3 + 1)(3 + 1) = 16$ .

STEP 9

Continue this process with other combinations of small primes, such as $2 \times 3 \times 5 = 30$ , and find the highest power of this product that is less than 999. This is $30^2 = 900$ .

STEP 10

Check the number of factors for $30^2$ , which is $2^2 \times 3^2 \times 5^2$ . The number of factors is $(2 + 1)(2 + 1)(2 + 1) = 27$ .

STEP 11

Since $30^2 = 900$ has more factors than the previous numbers we have checked, we will continue to look for numbers with more factors by considering other combinations of small primes and their powers.

STEP 12

Consider the combination of the first four primes $2 \times 3 \times 5 \times 7 = 210$ . The highest power of 210 that is less than 999 is $210^1 = 210$ .

STEP 13

Check the number of factors for $210$ , which is $2^1 \times 3^1 \times 5^1 \times 7^1$ . The number of factors is $(1 + 1)(1 + 1)(1 + 1)(1 + 1) = 16$ .

STEP 14

Since $210$ has fewer factors than $900$ , we can conclude that $900$ has more factors than any number that is a product of the first four primes.

STEP 15

Now consider numbers that are products of powers of the first three primes but with one of the primes raised to a higher power, such as $2^2 \times 3 \times 5 = 60$ , and find the highest power of this product that is less than 999. This is $60^2 = 3600$ , which is too large, so we consider $60^1 = 60$ .

STEP 16

Check the number of factors for $60$ , which is $2^2 \times 3^1 \times 5^1$ . The number of factors is $(2 + 1)(1 + 1)(1 + 1) = 12$ .

STEP 17

Since $60$ has fewer factors than $900$ , we can conclude that $900$ still has more factors than any number that is a product of the first three primes with one prime raised to the second power.

STEP 18

Continue to check other combinations of powers of small primes, ensuring that the product is less than 999.

STEP 19

After checking various combinations, we find that $2^3 \times 3^2 \times 5^1 = 360$ is a promising candidate.

STEP 20

Check the number of factors for $360$ , which is $2^3 \times 3^2 \times 5^1$ . The number of factors is $(3 + 1)(2 + 1)(1 + 1) = 24$ .

STEP 21

Since $360$ has fewer factors than $900$ , $900$ remains the number with the most factors we have found so far.

STEP 22

Conclude that after checking various combinations of small prime powers, $900$ has the most factors of any number between 9 and 999.
The number between 9 and 999 with the most factors is 900.

Was this helpful?

Find the integer between 9 and 999 with the most positive divisors.

STEP 1

STEP 2

Understand that the number of factors a number has is related to its prime factorization. The prime factorization of a number is the expression of the number as a product of its prime factors.

STEP 3

STEP 4

STEP 5

Start by considering the highest power of 2 that is less than 999, since 2 is the smallest prime number. This is 29=5122^9 = 51229=512.

STEP 6

Check the number of factors for 292^929. Since it has only one prime factor, the number of factors is 9+1=109 + 1 = 109+1=10.

STEP 7

Now consider the product of the smallest primes, such as 2×3=62 \times 3 = 62×3=6, and find the highest power of this product that is less than 999. This is 64=12966^4 = 129664=1296, which is greater than 999, so we need to consider 63=2166^3 = 21663=216.

STEP 8

Check the number of factors for 636^363, which is 23×332^3 \times 3^323×33. The number of factors is (3+1)(3+1)=16(3 + 1)(3 + 1) = 16(3+1)(3+1)=16.

STEP 9

Continue this process with other combinations of small primes, such as 2×3×5=302 \times 3 \times 5 = 302×3×5=30, and find the highest power of this product that is less than 999. This is 302=90030^2 = 900302=900.

STEP 10

Check the number of factors for 30230^2302, which is 22×32×522^2 \times 3^2 \times 5^222×32×52. The number of factors is (2+1)(2+1)(2+1)=27(2 + 1)(2 + 1)(2 + 1) = 27(2+1)(2+1)(2+1)=27.

STEP 11

Since 302=90030^2 = 900302=900 has more factors than the previous numbers we have checked, we will continue to look for numbers with more factors by considering other combinations of small primes and their powers.

STEP 12

Consider the combination of the first four primes 2×3×5×7=2102 \times 3 \times 5 \times 7 = 2102×3×5×7=210. The highest power of 210 that is less than 999 is 2101=210210^1 = 2102101=210.

STEP 13

Check the number of factors for 210210210, which is 21×31×51×712^1 \times 3^1 \times 5^1 \times 7^121×31×51×71. The number of factors is (1+1)(1+1)(1+1)(1+1)=16(1 + 1)(1 + 1)(1 + 1)(1 + 1) = 16(1+1)(1+1)(1+1)(1+1)=16.

STEP 14

Since 210210210 has fewer factors than 900900900, we can conclude that 900900900 has more factors than any number that is a product of the first four primes.

STEP 15

STEP 16

Check the number of factors for 606060, which is 22×31×512^2 \times 3^1 \times 5^122×31×51. The number of factors is (2+1)(1+1)(1+1)=12(2 + 1)(1 + 1)(1 + 1) = 12(2+1)(1+1)(1+1)=12.

STEP 17

Since 606060 has fewer factors than 900900900, we can conclude that 900900900 still has more factors than any number that is a product of the first three primes with one prime raised to the second power.

STEP 18

Continue to check other combinations of powers of small primes, ensuring that the product is less than 999.

STEP 19

After checking various combinations, we find that 23×32×51=3602^3 \times 3^2 \times 5^1 = 36023×32×51=360 is a promising candidate.

STEP 20

Check the number of factors for 360360360, which is 23×32×512^3 \times 3^2 \times 5^123×32×51. The number of factors is (3+1)(2+1)(1+1)=24(3 + 1)(2 + 1)(1 + 1) = 24(3+1)(2+1)(1+1)=24.

STEP 21

Since 360360360 has fewer factors than 900900900, 900900900 remains the number with the most factors we have found so far.

STEP 22

Conclude that after checking various combinations of small prime powers, 900900900 has the most factors of any number between 9 and 999.The number between 9 and 999 with the most factors is 900.

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

Start by considering the highest power of 2 that is less than 999, since 2 is the smallest prime number. This is $2^9 = 512$ .

Check the number of factors for $2^9$ . Since it has only one prime factor, the number of factors is $9 + 1 = 10$ .

Now consider the product of the smallest primes, such as $2 \times 3 = 6$ , and find the highest power of this product that is less than 999. This is $6^4 = 1296$ , which is greater than 999, so we need to consider $6^3 = 216$ .

Check the number of factors for $6^3$ , which is $2^3 \times 3^3$ . The number of factors is $(3 + 1)(3 + 1) = 16$ .

Continue this process with other combinations of small primes, such as $2 \times 3 \times 5 = 30$ , and find the highest power of this product that is less than 999. This is $30^2 = 900$ .

Check the number of factors for $30^2$ , which is $2^2 \times 3^2 \times 5^2$ . The number of factors is $(2 + 1)(2 + 1)(2 + 1) = 27$ .

Since $30^2 = 900$ has more factors than the previous numbers we have checked, we will continue to look for numbers with more factors by considering other combinations of small primes and their powers.

Consider the combination of the first four primes $2 \times 3 \times 5 \times 7 = 210$ . The highest power of 210 that is less than 999 is $210^1 = 210$ .

Check the number of factors for $210$ , which is $2^1 \times 3^1 \times 5^1 \times 7^1$ . The number of factors is $(1 + 1)(1 + 1)(1 + 1)(1 + 1) = 16$ .

Since $210$ has fewer factors than $900$ , we can conclude that $900$ has more factors than any number that is a product of the first four primes.

Check the number of factors for $60$ , which is $2^2 \times 3^1 \times 5^1$ . The number of factors is $(2 + 1)(1 + 1)(1 + 1) = 12$ .

Since $60$ has fewer factors than $900$ , we can conclude that $900$ still has more factors than any number that is a product of the first three primes with one prime raised to the second power.

After checking various combinations, we find that $2^3 \times 3^2 \times 5^1 = 360$ is a promising candidate.

Check the number of factors for $360$ , which is $2^3 \times 3^2 \times 5^1$ . The number of factors is $(3 + 1)(2 + 1)(1 + 1) = 24$ .

Since $360$ has fewer factors than $900$ , $900$ remains the number with the most factors we have found so far.

Conclude that after checking various combinations of small prime powers, $900$ has the most factors of any number between 9 and 999.
The number between 9 and 999 with the most factors is 900.