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.
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.
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:
From statement 2, we have:
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