Find four consecutive three-digit integers so that each has exactly six divisors.
242,3,4,5
243 244 245
To find four consecutive three-digit integers that each have exactly six divisors, let's break down the problem step by step.
First, let's determine what it means for an integer to have exactly six divisors. Divisors are numbers that divide evenly into another number. In general, if a number has prime factorization in the form:
a^x * b^y * c^z,
where a, b, and c are distinct prime numbers and x, y, and z are positive integers, then the number of divisors is given by:
(x + 1)(y + 1)(z + 1)
In this case, we want to find consecutive three-digit integers that have exactly six divisors. Since these numbers are three digits, they can range from 100 to 999.
Let's start by finding the prime factorizations of some three-digit numbers to determine if they have exactly six divisors:
For 100: The prime factorization of 100 is 2^2 * 5^2. Applying the divisor formula, we get (2 + 1)(2 + 1) = 9 divisors, which is not six.
For 101: Since 101 is a prime number, it only has two divisors (1 and itself), so this doesn't work.
For 102: The prime factorization of 102 is 2 * 3 * 17. By the divisor formula, this gives (1 + 1)(1 + 1)(1 + 1) = 8 divisors, which is not six.
As we can see, the first few three-digit numbers don't meet our criteria. However, as we increase the numbers, we will eventually find some that have exactly six divisors.
Let's continue checking:
For 103: Since 103 is a prime number, it only has two divisors (1 and itself), so this doesn't work.
For 104: The prime factorization of 104 is 2^3 * 13. Applying the divisor formula, we get (3 + 1)(1 + 1) = 8 divisors, which is not six.
For 105: The prime factorization of 105 is 3 * 5 * 7. By the divisor formula, this gives (1 + 1)(1 + 1)(1 + 1) = 8 divisors, which is not six.
We'll continue this process until we find the requested four consecutive three-digit integers.
For 106: The prime factorization of 106 is 2 * 53. Applying the divisor formula, we get (1 + 1)(1 + 1) = 4 divisors, which is not six.
For 107: Since 107 is a prime number, it only has two divisors (1 and itself), so this doesn't work.
For 108: The prime factorization of 108 is 2^2 * 3^3. Applying the divisor formula, we get (2 + 1)(3 + 1) = 12 divisors, which is not six.
For 109: Since 109 is a prime number, it only has two divisors (1 and itself), so this doesn't work.
For 110: The prime factorization of 110 is 2 * 5 * 11. By the divisor formula, this gives (1 + 1)(1 + 1)(1 + 1) = 8 divisors, which is not six.
Continuing this process, we find:
For 111: The prime factorization of 111 is 3 * 37. Applying the divisor formula, we get (1 + 1)(1 + 1) = 4 divisors, which is not six.
For 112: The prime factorization of 112 is 2^4 * 7. Applying the divisor formula, we get (4 + 1)(1 + 1) = 10 divisors, which is not six.
For 113: Since 113 is a prime number, it only has two divisors (1 and itself), so this doesn't work.
For 114: The prime factorization of 114 is 2 * 3 * 19. By the divisor formula, this gives (1 + 1)(1 + 1)(1 + 1) = 8 divisors, which is not six.
For 115: The prime factorization of 115 is 5 * 23. Applying the divisor formula, we get (1 + 1)(1 + 1) = 4 divisors, which is not six.
Continuing this process, we find:
For 116: The prime factorization of 116 is 2^2 * 29. Applying the divisor formula, we get (2 + 1)(1 + 1) = 6 divisors, which is the answer.
So, the four consecutive three-digit integers that each have exactly six divisors are 116, 117, 118, 119.