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
To solve this question, we need to find values for the unknown digits X and Y that make the fraction 5Y3X/2Y8 result in a quotient that is a whole number. Let's break down the steps to find the solution:
1. First, let's understand the given fraction: 5Y3X/2Y8
2. In order for the division to result in a whole number, the denominator (2Y8) must divide the numerator (5Y3X) evenly, with no remainder.
3. To understand the possible values of X, let's consider the divisibility rules. If the sum of the digits in a number is divisible by 3, then the number itself is also divisible by 3.
4. Looking at the denominator (2Y8), the sum of the digits is 2 + Y + 8 = 10 + Y. To make the denominator divisible by 3, the sum of the digits must be divisible by 3.
5. Now, let's examine the numerator (5Y3X). Since we want the division to result in a whole number, the numerator must also be divisible by the same factors as the denominator.
6. From step 4, we know that the denominator (2Y8) must be divisible by 3. Therefore, the numerator (5Y3X) must also be divisible by 3.
7. Let's consider the possible values of X: If X = 2, the numerator would evaluate to 5Y32. Now, check if this number is divisible by 3. If Y is a multiple of 3, then 5Y32 can be divisible by 3. However, if Y is not divisible by 3, then 5Y32 will not be divisible by 3. Hence, option 1, "X may equal 2" is true.
8. Now, let's consider the possible values of X: If X = 6 or 0, the numerator would be 5Y36 or 5Y30 respectively. Now, check if these numbers are divisible by 3. Similar to step 7, if Y is a multiple of 3, then these numbers can be divisible by 3. However, if Y is not divisible by 3, then 5Y36 and 5Y30 will not be divisible by 3. Hence, option 11, "X may equal 6 or 0" is true.
9. Finally, let's consider the possible values of X: If X = 4, the numerator would be 5Y34. Now, check if this number is divisible by 3. Again, if Y is a multiple of 3, then the number can be divisible by 3. However, if Y is not divisible by 3, then 5Y34 will not be divisible by 3. Hence, option 111, "X may equal 4" is not true.
Based on the above analysis, the correct answer is option B.11 only, which means that X may equal 6 or 0.