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Iteru [2.4K]
4 years ago
15

Here is another for fun question. MUST SHOW ALL WORKINGS

Mathematics
1 answer:
romanna [79]4 years ago
3 0
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 \boxed{8521}
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#CarryOnLearning

5 0
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The GCF of 12, 21, and 30 is 3.
The factors of 12 are: 1, 2, 3, 4, 6, 12
<span>The factors of 21 are: 1, </span>3, 7, 21
<span>The factors of 30 are: 1, 2, </span>3, 5, 6, 10, 15, 30
<span>Then the greatest common factor is 3.
</span>
The GCF of 11, and 44 is 11. 
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The factors of 44 are: 1, 2, 4, 11, 44
Then the greatest common factor is 11. 

Hope this helps.
8 0
3 years ago
Read 2 more answers
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