math
posted by vincent on .
The fraction 5Y3X/2Y8, in which x and Y stand for two unknown digits, represents a division which results in a quotient that is a whole number. Which of the following is (are) true?
1. X may equal 2.
11. X may equal 6 or 0.
111. X may equal 4.
not just and answer, i don't get this question very much.
A.1 only
B.11 only
C.111 only
D.1 and 111 only
E.1,11,and 111

3X5Y/2Y8 = 15x/16.
15x/16 = 1 = The smallest whole number.
15x = 16
X = 16/15 = 1 1/15.
According to my analysis, x cannot equal
0,2,4, nor 6. 
f y=0 > 503?/208 = appr 24
to be exact 24*208 = 4992 , so y ≠ 0
if y = 1 > 513?/218 = appr 24
24*218 = 5232 , so y≠1
y=2 > 523?/228 = 23
23*228 = 5244 , so y≠2
y=3 > 533?/238 = appr 22
22*238 = 5236 , so y≠3
y = 4 > 543?/248 = appr 22
22*248 = 5456 , so y≠4
y = 5 > 553?/258 = appr 21
21*258 = 5418 , so y≠5
y = 6 > 563?/268 = appr 21
21*268 = 5628
20*268 = 5360 , y ≠ 6
y = 7 > 573?/278 = appr 21
21*278 = 5838
20*278 = 5560 , y ≠ 7
y = 8 > 583?/288 = appr 20
20*288 = 5760
21*288 = 6048
19*288 = 5472 , y ≠ 8
y = 9 > 593?/298 = appr 21
21*298 = 6258
20*298 = 5960
22*298 = 6556
I have exhausted all possible choices for y from 0 to 9
since x depends on what y values we choose,
and unless I made an arithmetic error, no values of x and y make the statement true. 
I. X may equal 2.
II. X may equal 6 or 0.
III. X may equal 4.
A.I only
B.II only
C.III only
D.I and III only
E.I,II,and III