The 2-digit positive integer x has the property that it is divisible

Stat 1) Cyclic power of 3 and 7 will have 7 as a unit digit

Any two digit number of the form \(A3^3\) has unit digit 7 and \(A7^1\) will have unit digit 7

Now for the given condition that The 2-digit positive integer x has the property that it is divisible by its units digit we can multiple values of X satisfying statement 1
for eg 33 , 63, 93, 77 ; Multiple values of X hence Not suff

Stat 2) X+1 is also divisible by its unit digit
here we can also have multiple values of X for eg
X = 32 , 63
X+1 = 33, 64
hence NOt Suff


Together

We can have only one value ie 63 that satisfies both condition hence Suff Ans C

From 1: Since X^3 has a unit digit 7, therefore, unit digit of x must be 3.
Now two digit numbers, which are divisible to its own unit digit are 33,63 & 93 - Since Multiple answer, therefore not sufficient

From 2: if x=11 then x+1=12, all of these are divisible by it's unit digit. Similarly, (21,22), (31,32)......- Multiple answer- Not Sufficient

Combining 1 & 2, we have only 63, which satisfy both.

Hence, C is the answer.

Hi,

I have understand why and how a & b independently are not sufficient but can't seem to understand how both of them together are sufficient. i.e. how does 63 satisfy both the equations.
63+1=64 which is not divisible by 3 the units digit of 63.
Please let me know.

it is not about 64/3, instead it is 64/4 [ unit digit of the number]

mrinalsharma1990
It's not 63 rather 63+1=64 is divisible by 4.

Hope it helps!

How we got number 33, 63, 93 and 77?

Sent from my RNE-L21 using GMAT Club Forum mobile app

X is div by its units digit..

For some info on property

a) All numbers finishing with 1 for example.. 11,21,..91
b) All numbers finishing with 2 for example.. 22,32..92
c) Numbers with units digit 3 and tens digit multiple of 3.. 33,63,93
d) Numbers with units digit 4 and even digit in tens... 24,44,64,84
e) All numbers ending with 5... 15,25,...95
f) Numbers with units digit 6 and multiple of 3 in tens digit..36,66,96
g) Numbers ending with 8 and even non multiple of 4 in tens digit... 24,64
h) Numbers ending with 9 and multiple of 9 in tens..99

1) x^3 has a units digit of 7..
Only 3 fits in as per cyclicity..
So X has units digit 3, but all numbers 33,63,93 as shown above fit in..
Insufficient

2) X+1 is also div by its units digit.
As can be seen all numbers ending with 1 will become even and div by 2, the new units digit.
Also 63to 64..
Insufficient

Combined..
Statement I tells those ending with 3 and statement II says out of these only 63 fits in..
Sufficient

C
_________________

The 2-digit positive integer x has the property that it is divisible by its units digit. What is x?

(1) x^3 has a units digit of 5

(2) x+1 is also divisible by its units digit.

Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient to answer the question asked
Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient to answer the question asked
Both statements (1) and (2) TOGETHER are sufficient to answer the question asked; but NEITHER statement ALONE is sufficient
EACH statement ALONE is sufficient to answer the question asked
Statements (1) and (2) TOGETHER are NOT sufficient to answer the question asked, and additional data specific to the problem are needed

Statement 1 suggests that X^3 has unit digit of 5. Hence we know the unit digit of number is 5. as only 5^3 will have the number 5 as unit digit.
So the possible two digit number can be 15,25,35,45,55,65,75,85,95.
statement 1 is not sufficient.

Statement 2 says x+1 is divisible by its own unit digit.
We can have number pairs of X and X+1 as 11 and 12, 21 and 22, 31 and 32.
So clearly not sufficient.

Statement 1 and statement 2 combined will leave us with two numbers 35,65 both have x+1 divisible by it unit digit.
Hence not sufficient.

So answer is E.

Please give kudos if you like the explanation.

_______________
Merging topics.
_________________

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________