If the units digit of integer n is greater than 2, what is

Thank you. I also thought that it was not a hard problem. Though I'll wait for other comments and then might change difficulty to 600.

Hi,

Difficulty level: 600

Since, unit digit is greater than 2,
then, possible digits at unit place would be:
3: Unit digit of \(3^2=9, 3^3=7\)
4: Unit digit of \(4^2=6, 4^3=4\)
5: Unit digit of \(5^2=5, 5^3=5\)
6: Unit digit of \(6^2=6, 6^3=6\)
7: Unit digit of \(7^2=9, 7^3=3\)
8: Unit digit of \(8^2=4, 8^3=2\)
9: Unit digit of \(9^2=1, 9^3=9\)

Using (1),
Unit digit can be:
5: Unit digit of \(5^2=5\)
6: Unit digit of \(6^2=6\). Insufficient.

Using (2),
Unit digit can be:
4: Unit digit of \(4^3=4\)
5: Unit digit of \(5^3=5\)
6: Unit digit of \(6^3=6\). Insufficient.

Combining both the statements:
Unit digit can either be 5 or 6. Insufficient.

Thus, Answer is (E)

Regards,

Difficulty level: 600+

I simply noted down numbers from 3 to 9 and checked square and cube values. Found 5 and 6 given similar answers.

E for me

If the units digit of integer n is greater than 2, what is the units digit of n ?

(1) The units digit of n is the same as the units digit of n^2.
n=5,6 both greater than 2
5^2=25
6^2=36
not sufficient
(2) The units digit of n is the same as the units digit of n^3.
n=4,5,6,9
4^3=64
5^3=125
6^3=216
9^3=729
not sufficient
from (i) and (ii)
5 & 6 not sufficient
(E)

Here is my solution ->
Units digit of n>2
so the unit digit of n can be => {3,4,...9}
we need the unit digit of n
Lets dive into statements
Statement 1
Unit digit of n = units digit of n^2
so the units digits of n can be 5 or 6
Not sufficient
Statement 2
Units digit of n =units digit of n^3
so the units digit of n^3 can be {4,5,6,9}
Hence not sufficient
Statement 1 and statement 2
units digit of n can be 5 or 6
Hence insufficient
Hence E

Theory:
unit-digit^p = unit-digit, when unit-digit is 0,1,5,6. p may be odd or even

(--4)^odd=4
(--9)^odd=9

Statement 1.
5,6 (not, 4 and 9 because the power is square(even)). Insufficient

statement 2.
5,6, and 4,9. Insufficient

Statement 1 and 2.
5,6. Two values so insufficient

Hence, Ans is E

We are given that the units digit of integer n is greater than 2 and we need to determine that units digit.

Statement One Alone:

The units digit of n is the same as the units digit of n^2.

If the units digit of n is the same as the units digit of n^2, then the units digit can be 0, 1, 5, or 6 (since 0^2 = 0, 1^2 = 1, 5^2 = 25 and 6^2 = 36). However, we are given that the units digits is greater than 2, so n can be either 5 or 6. Since we can’t determine exactly which digit it is, statement one alone is not sufficient to answer the question.

Statement Two Alone:

The units digit of n is the same as the units digit of n^3.

If the units digit of n is the same as the units digit of n^2, then the units digit can be 0, 1, 4, 5, 6, or 9. (since 0^3 = 0, 1^3 = 1, 4^3 = 64, 5^3 = 125, 6^3 = 216 and 9^3 = 729). However, we are given that the units digits is greater than 2, so n can be 4, 5, 6 or 9. Since we can’t determine exactly which digit it is, statement two alone is not sufficient to answer the question.

Statements One and Two Together:

With two statements together, the units digit of n can still be either 5 or 6. Since we can’t determine which one it is, the two statements together are still not sufficient to answer the question.

Answer: E

So, n can be 3, 4 ,5,6,7,8 and 9.

So let's find squares and cubes of these numbers.

3^2 = 9 so last digit is not 3. Similarly for 4 , 7,8 and 9, when we square or cube them units digit changes. For 5 and 6 units digit doesn't not change for square or cube. So we have 2 choices. Don't know which one is the one.

So the answer is E

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For the given statement n ; Lets xy be the number where y>2 and we need n=?

1. n unit digit = n^2 unit digit
5 , 25 and 6,36 -- not sufficient
2. n unit digit = n^3 unit digit
5,125 , 6,216 rest we dont need. -- not sufficient

we have taken example above which shows we still don't have one value combined

Ans E.

Nice one, I think we can bump this to 600+

n= An integer, say for the sake of simplicity:

\(n= _ _\)

(1) The units digit of n is the same as the units' digit of \(n^2\).

let say, \(n= 6\), we can write 6 as two-digit integer = \(06\) and \(n^2 = 36\) ( Same units digit 6)

let say, \(n= 5\), then \(n^2= 25\) (say units digit of 5)

we have two possible values, not sufficient.

(2) The units digit of n is the same as the units' digit of \(n^3\).

again, let's say \(n= 06\), \(n^3= 216\) ( same units digit, 6)

Say, \(n=9\) , \(n^3= 729\) (same units digit, 9)

Two possible values, insufficient.

1+2

We can re-use the same values, for simplicity,

n has the same values for Unit digit as n^2 and n^3, Cool : what if n= 6 and n= 11...? both \(n^2 \) and \(n^3 \)for 11 and 6 will have the same unit's digit. hence, we can't find n based on the provided info.

E. is the answer

"since the units digit of n is greater than 2, then its units digit could be 5 or 6"

but the units digit could also be 7 and 9, e.g.: 7, 7'2=49

What don't I understand here?

"since the units digit of n is greater than 2, then its units digit could be 5 or 6"

but the units digit could also be 7 and 9, e.g.: 7, 7'2=49

What don't I understand here?


Okay I got it, unit digits of n has to equal units digit of n'2...

(1) Why can the units digit only be 5 or 6? 7'2=49, so the units digit here is 9

(1) says: The units digit of n is the same as the units digit of n^2. Does ..7 or ..9 satisfy this?