Here, we need to find the GCF of color(orange)12 and color(green)812812 . So lets list the factors:
color(red)(12): 1, 2, 3, 4, 6, 12
color(blue)(812812) : 1, 2, 4, 7, 11, ...
From here, we can see that they both have color(magenta)4. So you can make 4 batches of trail mix.
- If find the factors of 812812 is difficult, here's another way:
color(green)1. Look at the factors of color(orange)12: 1, 2, 3, 4, 6, 12
color(green)2.Using the divisibility rules, see if color(red)(2,3,4,6), or color(red)12 is one of color(blue)812812's factors.
color(green)3.From just looking at 812812, we can see that color(red)12 and color(gold)812812 both have color(magenta)2 as a common factor. ** * Because to see if a number is divisible by color(magenta)2, the rule is that the last digit of the number is even. * **
color(green)4. To see if color(lightgreen)812812 is divisible by color(pink)3, let's use the divisibility rule for 3. Which is : ** * add the digits of the number up and see if it's divisible by 3. * ** (color(purple)(8+1+2+8+1+2) = 22. Thus, color(lightgreen)812812 is not divisible by color(pink)3
color(green)5. Let's now try color(red)4. The divisibility rule for color(red)4 is if the last two digits are divisible by color(red)4, * or if the last two digits are zeros . Now take a look at color(blue)812812. color(green)812812's last two digits is 12. We know that 12 is divisible by color(red)4, so that means color(blue)812812 is divisible by color(red)4.
color(green)6. Continue the steps above for color(red)(6 , 12)
color(red)( . . . )
In the end, we see that color(red)4 is the GCF of color(orange)12 and color(lightgreen)812812 . So you can make color(red)4 batches.
