What is the last two digits of m when m = 10^n - 32?

What is the last two digits of m when m = 10^n - 32?

• A. 32
• B. 68
• C. 96
• D. 00

Answer: (B)

To determine the last two digits of \( m \) when \( m = 10^n - 32 \), we need to evaluate \( 10^n \mod 100 \) since the last two digits of a number can be derived from its value modulo 100. For any integer \( n \geq 2 \), \( 10^n \) is clearly divisible by 100, which means that \( 10^n \equiv 0 \mod 100 \). Thus, we have: \[ m = 10^n - 32 \equiv 0 - 32 \mod 100 \equiv -32 \mod 100. \] To convert \(-32\) into a positive representation under modulo 100, we can add 100: \[ -32 + 100 = 68. \] Therefore, the last two digits of \( m \), when \( n \geq 2 \), are 68. Hence, the correct answer is that the last two digits of \( m \) are indeed 68. This solution gives a clear understanding of how modular arithmetic can be applied to find the last two digits of a larger expression.

To determine the last two digits of ( m ) when ( m = 10^n - 32 ), we need to evaluate ( 10^n \mod 100 ) since the last two digits of a number can be derived from its value modulo 100.

For any integer ( n \geq 2 ), ( 10^n ) is clearly divisible by 100, which means that ( 10^n \equiv 0 \mod 100 ). Thus, we have:

[

m = 10^n - 32 \equiv 0 - 32 \mod 100 \equiv -32 \mod 100.

]

To convert (-32) into a positive representation under modulo 100, we can add 100:

[

-32 + 100 = 68.

]

Therefore, the last two digits of ( m ), when ( n \geq 2 ), are 68. Hence, the correct answer is that the last two digits of ( m ) are indeed 68. This solution gives a clear understanding of how modular arithmetic can be applied to find the last two digits of a larger expression.