At Westonci.ca, we make it easy to get the answers you need from a community of informed and experienced contributors. Discover the answers you need from a community of experts ready to help you with their knowledge and experience in various fields. Explore comprehensive solutions to your questions from a wide range of professionals on our user-friendly platform.
Sagot :
For the product of the digits to be divisible by 10, it must have 10 as a factor, of course. Since 10 can't act as a digit, this means two of our digits must be 2 & 5.
No digit can be 0 because this would give a total product of 0.
Neither 2 nor 5 can be our largest because the lowest you can get is 1+2+3=6.
The lowest possible largest digit would have to be 1+2+5=8.
As for 9, this is not an option, and here's why: We need that 2 and 5, but to get 9 our other digit would have to be 2, which is a repeat.
So, our digits are 1, 2, 5, and 8.
It doesn't matter which order they are in. (take a look at the rules)
Therefore, we should put the highest digits in the highest value places.
The answer is [tex]\boxed{8521}[/tex]
No digit can be 0 because this would give a total product of 0.
Neither 2 nor 5 can be our largest because the lowest you can get is 1+2+3=6.
The lowest possible largest digit would have to be 1+2+5=8.
As for 9, this is not an option, and here's why: We need that 2 and 5, but to get 9 our other digit would have to be 2, which is a repeat.
So, our digits are 1, 2, 5, and 8.
It doesn't matter which order they are in. (take a look at the rules)
Therefore, we should put the highest digits in the highest value places.
The answer is [tex]\boxed{8521}[/tex]
We hope this was helpful. Please come back whenever you need more information or answers to your queries. We appreciate your visit. Our platform is always here to offer accurate and reliable answers. Return anytime. Westonci.ca is your trusted source for answers. Visit us again to find more information on diverse topics.