Around the World in 80 Questions (Day 8): If d is a single-digit numb

8 possible values of d = 2,3,4,5......,8,9

(1) 2 is not a factor of 23
3 is a factor of 33
4 is not a factor of 43
5 is not a factor of 53
6 is not a factor of 63
7 is not a factor of 73
8 is not a factor of 83
So d= 3---------SUFFICIENT

(2) 2 is not a factor of 29
3 is a factor of 39
4 is not a factor of 49
5 is not a factor of 59
6 is not a factor of 69
7 is not a factor of 79
8 is not a factor of 89
9 is a factor of 99

So d can be 3 or 9.-----------> NOT SUFFICIENT

OA: A

If d is a single-digit number greater than 1, what is the value of d ?

(1) d is a factor of the two digit number d3, where d is the tens digit and 3 is the units digit.
(1) d is a factor of the two digit number d9, where d is the tens digit and 9 is the units digit.

Given d >1 and is a single-digit number
d?

Case I: d is a factor of the two digit number d3
d can be 1 as 1 is a factor of 13, but d >1
d can be 3 as 3 is a factor of 33
so we get only one value of d here as 3
Hence this is sufficient

Case II: d is a factor of the two digit number d9
d can be 1 as 1 is a factor of 19, but given that d >1 this choice is void
d can be 3 as 3 is a factor of 39
d can be 9 as 9 is a factor of 99
Here we get two values of d
hence this is not sufficient

A is the answer

Given: If d is a single-digit number greater than 1, what is the value of d ?

(1) d is a factor of the two digit number d3, where d is the tens digit and 3 is the units digit.
(1) d is a factor of the two digit number d9, where d is the tens digit and 9 is the units digit.

Solution:
d > 1 and a single digit number
That means 1 < d <=9

Option 1: d is a factor of the two digit number d3, where d is the tens digit and 3 is the units digit.

2-digit numbers in the form of d3 are 23, 33, 43, 53, 63, 73, 83, 93 (13 isn't possible as d > 1)

Now if d = 2, 2 is not a factor of 23
if d = 3, 3 is a factor of 33
if d = 4, 4 is not a factor of 43
if d = 5, 5 is not a factor of 53
if d = 6, 6 is not a factor of 63
if d = 7, 7 is not a factor of 73
if d = 8, 8 is not a factor of 83
if d = 9, 9 is not a factor of 93

We get only 1 possible value of d3 i.e. 33
so that d = 3

SUFFICIENT

Option 2: d is a factor of the two digit number d9, where d is the tens digit and 9 is the units digit.

2-digit numbers in the form of d9 are 29, 39, 49, 59, 69, 79, 89, 99 (19 isn't possible as d > 1)

Now if d = 2, 2 is not a factor of 29
if d = 3, 3 is a factor of 39
if d = 4, 4 is not a factor of 49
if d = 5, 5 is not a factor of 59
if d = 6, 6 is not a factor of 69
if d = 7, 7 is not a factor of 79
if d = 8, 8 is not a factor of 89
if d = 9, 9 is a factor of 99

We get 2 possible value of d9 i.e. 39 and 99
so that d = 3 or d = 9

INSUFFICIENT

Option A is the correct choice

1<\(d\leq{9}\)


d3 is an odd number.
d = 3 or 5 or 7 or 9 (Even number cannot be a factor of an odd number)
Since last digit is 3. Therefor \(d\neq{5}\).

Testing cases:
A. d = 3; 33 has 3 as one of its factor. (This case is valid)
B. d =7; 73 doesn't have 7 as one of its factor. (Invalid case)
C. d = 9; 93 doesn't have 9 as one of its factor. (Invalid case)

=> d =3. Statement 1 is SUFF.


d9 is an odd number.
d = 3 or 5 or 7 or 9 (Even number cannot be a factor of an odd number)
Since last digit is 3. Therefor \(d\neq{5}\).

Testing cases:
A. d = 3; 39 has 3 as one of its factor. (This case is valid)
B. d =7; 79 doesn't have 7 as one of its factor. (Invalid case)
C. d = 9; 99 has 9 as one of its factor. (Valid case)

=> d =3 or 9. Statement 2 is INSUFF.

IMO OA should be A.