C ,fam im on the cite and everythanng,
34? Or 32? I’m not sure cause it’s a 2 step problem and math isn’t the best thing I’m good at
let's firstly convert the mixed fractions to improper fractions, and then subtract.
![\bf \stackrel{mixed}{10\frac{1}{3}}\implies \cfrac{10\cdot 3+1}{3}\implies \stackrel{improper}{\cfrac{31}{3}}~\hfill \stackrel{mixed}{13\frac{1}{2}}\implies \cfrac{13\cdot 2+1}{2}\implies \stackrel{improper}{\cfrac{27}{2}} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \cfrac{27}{2}-\cfrac{31}{3}\implies \stackrel{\textit{using the LCD of 6}}{\cfrac{(3)27~~-~~(2)31}{6}}\implies \cfrac{81~~-~~62}{6}\implies \cfrac{19}{6}\implies 3\frac{1}{6}](https://tex.z-dn.net/?f=%5Cbf%20%5Cstackrel%7Bmixed%7D%7B10%5Cfrac%7B1%7D%7B3%7D%7D%5Cimplies%20%5Ccfrac%7B10%5Ccdot%203%2B1%7D%7B3%7D%5Cimplies%20%5Cstackrel%7Bimproper%7D%7B%5Ccfrac%7B31%7D%7B3%7D%7D~%5Chfill%20%5Cstackrel%7Bmixed%7D%7B13%5Cfrac%7B1%7D%7B2%7D%7D%5Cimplies%20%5Ccfrac%7B13%5Ccdot%202%2B1%7D%7B2%7D%5Cimplies%20%5Cstackrel%7Bimproper%7D%7B%5Ccfrac%7B27%7D%7B2%7D%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%20%5Ccfrac%7B27%7D%7B2%7D-%5Ccfrac%7B31%7D%7B3%7D%5Cimplies%20%5Cstackrel%7B%5Ctextit%7Busing%20the%20LCD%20of%206%7D%7D%7B%5Ccfrac%7B%283%2927~~-~~%282%2931%7D%7B6%7D%7D%5Cimplies%20%5Ccfrac%7B81~~-~~62%7D%7B6%7D%5Cimplies%20%5Ccfrac%7B19%7D%7B6%7D%5Cimplies%203%5Cfrac%7B1%7D%7B6%7D)
The only way for two integers to have an odd product is each integer is odd.
For example, 1*3=3 (odd), 1 and 3 are both odd.
Or,
5*11=55, all of 5,11,55 are odd.
The sum of two odd integers is always even, so the condition of even sum is automatically satisfied when the product is odd.
Out of the four integers, there are only two odd numbers, so choose the pair to be these two odd numbers and you'd get the right answer.
