The positive number X is divisible by 42, and is composed of only 1s and 0s when written in base 10. What's the smallest number that X might be?
clearly, x must be a multiple of 5*42 = 210
I get 101010 = 42*2405 = 210*481
To find the smallest number that meets the given conditions, we need to determine the properties of the number.
We are given that the number is divisible by 42. To find the divisors of 42, we can calculate its prime factorization: 42 = 2 x 3 x 7.
To determine if a number is divisible by another number, all the prime factors of the divisor must also be prime factors of the number. Therefore, the number in question must have 2, 3, and 7 as prime factors.
Next, we are given that the number is composed only of 1s and 0s when written in base 10. This means that the number is a binary number, consisting of only 0s and 1s.
To satisfy the condition that the number has 2, 3, and 7 as prime factors, we need to ensure that those numbers can be expressed in binary.
In binary, 2 is represented as 10, 3 is represented as 11, and 7 is represented as 111.
To determine the smallest number that satisfies the given conditions, we need to find the least common multiple (LCM) of 10, 11, and 111. The LCM is the smallest number that is divisible by all three numbers.
Calculating the LCM of 10, 11, and 111, we get:
LCM(10, 11, 111) = 1110.
Therefore, the smallest number, X, that satisfies the given conditions is 1110.