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.
