All the products of a multiplication fact with ___ as a factor have the digits 0 and 5 in the ones place

Question

Is the answer 2 or 5 or 10

Answer 1

The correct answer is 10.

When a multiplication fact includes 10 as a factor, the products will always end with the digits 0 and 5. This is due to the fact that multiplying any number by 10 results in a number that has 0 in the ones place.

If you multiply 10 by an even number, it would end in 0 (e.g. 10 x 2 = 20), and if you multiply 10 by an odd number, it would also end in 0 (e.g. 10 x 3 = 30, 10 x 5 = 50). However, it should be noted that only when observing the multiples of 5 specifically, do you find products ending in 0 and 5. But with 10 as a factor, 0 is dominant since the last digit will always be 0.

In summary, among the options given (2, 5, or 10), the only factor that guarantees a product that can include both 0 and 5 in the ones place consistently is 10.

Answer 2

wrong. A factor of 10 guarantees a 0 in the 1's place

Answer 3

You're correct, I appreciate your clarification. A factor of 10 guarantees a 0 in the ones place.

If we're looking for a factor that produces products ending in both 0 and 5, the factor must be 5.

When you multiply any number by 5, the resulting product can end in either 5 (if the other factor is odd) or 0 (if the other factor is even):

• For example:
  • \(5 \times 1 = 5\) (ends in 5)
  • \(5 \times 2 = 10\) (ends in 0)
  • \(5 \times 3 = 15\) (ends in 5)
  • \(5 \times 4 = 20\) (ends in 0)

Therefore, the answer to your question is 5, as it is the factor that allows the products to have digits 0 and 5 in the ones place. Thank you for your patience!