<span>33 =1, 3, 11, 33
75</span><span>=1, 3, 5, 15, 25, 75</span>
first off, let's convert the mixed fraction to improper fraction and then proceed, let's notice that by PEMDAS or order of operations, the multiplication is done first, and then any sums.
![\stackrel{mixed}{1\frac{7}{8}}\implies \cfrac{1\cdot 8+7}{8}\implies \stackrel{improper}{\cfrac{15}{8}} \\\\[-0.35em] ~\dotfill\\\\ -\cfrac{3}{4}~~ + ~~\cfrac{15}{8} \div \cfrac{1}{2}\implies -\cfrac{3}{4}~~ + ~~\cfrac{15}{8} \cdot \cfrac{2}{1}\implies -\cfrac{3}{4}~~ + ~~\cfrac{15}{4} \\\\\\ \cfrac{-3+15}{4}\implies \cfrac{12}{4}\implies 3](https://tex.z-dn.net/?f=%5Cstackrel%7Bmixed%7D%7B1%5Cfrac%7B7%7D%7B8%7D%7D%5Cimplies%20%5Ccfrac%7B1%5Ccdot%208%2B7%7D%7B8%7D%5Cimplies%20%5Cstackrel%7Bimproper%7D%7B%5Ccfrac%7B15%7D%7B8%7D%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20-%5Ccfrac%7B3%7D%7B4%7D~~%20%2B%20~~%5Ccfrac%7B15%7D%7B8%7D%20%5Cdiv%20%5Ccfrac%7B1%7D%7B2%7D%5Cimplies%20-%5Ccfrac%7B3%7D%7B4%7D~~%20%2B%20~~%5Ccfrac%7B15%7D%7B8%7D%20%5Ccdot%20%5Ccfrac%7B2%7D%7B1%7D%5Cimplies%20-%5Ccfrac%7B3%7D%7B4%7D~~%20%2B%20~~%5Ccfrac%7B15%7D%7B4%7D%20%5C%5C%5C%5C%5C%5C%20%5Ccfrac%7B-3%2B15%7D%7B4%7D%5Cimplies%20%5Ccfrac%7B12%7D%7B4%7D%5Cimplies%203)
Answer:
Step-by-step explanation:
To prove divisibility, we need to factor the divident such that one of its factors matches the divisor.
(I use the notation x|y to denote that x divides y)
(A)

(B)

In this case, it is easier to also factor the divisor to primes:

Both of these factor must be matched in the dividend in order to prove divisibility, and that indeed turns out to be true:
