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If n is an integer such that 3 <= n <= 7, what is the unit’s digit of

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If n is an integer such that 3 <= n <= 7, what is the unit’s digit of  [#permalink]

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New post 02 Apr 2018, 21:20
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A
B
C
D
E

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  35% (medium)

Question Stats:

80% (01:32) correct 20% (01:43) wrong based on 57 sessions

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Location: India
Concentration: General Management, Marketing
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Re: If n is an integer such that 3 <= n <= 7, what is the unit’s digit of  [#permalink]

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New post 02 Apr 2018, 21:29
Bunuel wrote:
If n is an integer such that 3 <= n <= 7, what is the unit’s digit of n^2?

(1) (n + 1)^2 has units digit 6.
(2) (n - 1)^2 has units digit 6.



n can be 3,4,5,6,7

From1:n can be 3 and 5 ,
for n=3,N+1=3+1=4 ,4^2 unit digit =6
similarly for 5 unit digit will be 6.
insufficient

from2:n can be 5 and 9
Insufficient

combining both n=5

hence C
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Re: If n is an integer such that 3 <= n <= 7, what is the unit’s digit of  [#permalink]

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New post 02 Apr 2018, 23:11
Bunuel wrote:
If n is an integer such that 3 <= n <= 7, what is the unit’s digit of n^2?

(1) (n + 1)^2 has units digit 6.
(2) (n - 1)^2 has units digit 6.


As there are only a small number of options, we'll just write them out.
This is an Alternative approach.

2^2 = 4
3^2 = 9
4^2 = 16
5^2 = 15
6^2 = 36
7^2 = 49
8^2 = 64

(1) is true for 4 and 6 meaning n=3 or 5. In this case the answer to 'what is the units digit of n^2' is '9' or '5'.
Insufficient.

(2) is true for 4 or 6 meaning n=5 or 7. In this case the answer to our question is '5' or '9'.
Insufficient.

Combined.
Combining the above means that n = 5 so our answer is '5'.
Sufficient.

(C) is our answer.
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Re: If n is an integer such that 3 <= n <= 7, what is the unit’s digit of  [#permalink]

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New post 03 Apr 2018, 04:20

Solution



Given:
    • n is an integer such that 3<=n<=7.

To find:
    • What is the units digit of \(n^2\)?

Statement-1:\((n + 1)^2\) has units digit 6.“

Since 3<=n<=7, 4<=(n+1) <=8

The units digit of all the possible values of \((n+1)^2\) is:
    • \(4^2= 6\)
    • \(5^2=5\)
    • \(6^2=6\)
    • \(7^2=9\)
    • \(8^2=4\)

For (n+1) = 4 and (n+1)=6, the units digit of \((n+1)^2\) is 6.
    • Thus, n= 3 or 5.

We do not have a unique value of n. Thus, Statement 1 alone is not sufficient to answer the question.

Statement-2:\((n - 1)^2\) has units digit 6.“

Since 3<=n<=7, 2<=(n+1) <=6

The units digit of all the possible values of \((n-1)^2\) is:
    • \(2^2= 4\)
    • \(3^2=9\)
    • \(4^2=6\)
    • \(5^2=5\)
    • \(6^2=6\)

For (n-1) = 4 and (n-1)=6, the units digit of \((n-1)^2\) is 6.
    • Thus, n= 5 or 7.

We do not have a unique value of n. Thus, Statement 2 alone is not sufficient to answer the question.

Combining both the statements:

From statement 1, we have:
    • n= 3 or 5

From statement 2, we have:
    • n= 5 or 7

By combining both the statements, the common value of n is 3 and the units digit of \(n^2\) is 9.

Hence, Statement (1) and (2) together are sufficient to answer the question.

Hence, the correct answer is option C.

Answer: C

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