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

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

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.

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