Final answer:
Jake's observation is correct because multiples of both 6 and 7 are also multiples of 42, and the ones digit in multiples of 42 is determined by the even number 2, thus resulting in an even digit in the ones place.
Step-by-step explanation:
Jake stated that any number which is a multiple of both 6 and 7 must have an even digit in the ones place. To understand this, let's explore the multipliers of 6 and 7 separately. Firstly, the multipliers of 6 (like 6, 12, 18, 24, ...) always end in an even digit because 6 itself is an even number and even numbers multiplied together will always result in an even number.
Similarly, the multipliers of 7 (e.g., 7, 14, 21, 28, ...) alternate between ending with an odd digit and an even digit. However, to be a multiple of both 6 and 7, a number must be a multiple of their least common multiple, which is 42. Multiples of 42 (42, 84, 126, 168, ...) exhibit the pattern that they all end in an even digit.
This occurs because when you multiply 42 by any integer, the last digit of the result will be determined by the last digit of 42, which is 2, an even number. Therefore, any number that is a multiple of both 6 and 7 will indeed have an even digit in the ones place.