$4^{2^{40}}-3^{01^2}

The last digit of the expression 4 2 40 − 3 0 1 2 simplifies to the last digit of 2 2 41 − 3 . Since the last digit of 2 2 41 is 6, subtracting 3 gives a final result of 3. Thus, the last digit of the given expression is 3.
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Answered by Anonymous | 2025-07-04

Simplify the expression: 4 2 40 − 3 0 1 2 = 4 2 40 − 3 .
Rewrite the expression: 4 2 40 − 3 = 2 2 41 − 3 .
Find the last digit of 2 2 41 by considering the cycle of last digits of powers of 2 and finding 2 41 ( mod 4 ) , which is 0. Thus, the last digit of 2 2 41 is 6.
Calculate the last digit of the final expression: 6 − 3 = 3 . Therefore, the last digit of 4 2 40 − 3 0 1 2 is 3 ​ .

Explanation

Simplifying the Expression We are asked to evaluate the expression 4 2 40 − 3 0 1 2 . First, we simplify the expression. Note that 0 1 2 = 1 , so the expression becomes 4 2 40 − 3 1 or 4 2 40 − 3 .

Rewriting the Expression We can rewrite the expression as ( 2 2 ) 2 40 − 3 = 2 2 ⋅ 2 40 − 3 = 2 2 41 − 3 . The problem asks to evaluate the expression, which likely means to find the last digit or some other property of the number.

Finding the Last Digit Consider finding the last digit of 2 2 41 . The last digits of powers of 2 cycle with a period of 4: 2, 4, 8, 6, 2, 4, 8, 6, ... We need to find the exponent 2 41 ( mod 4 ) .

Calculating the Exponent Modulo 4 Since 2 41 is divisible by 4 (as 2 41 = 4 ⋅ 2 39 ), 2 41 ≡ 0 ( mod 4 ) . Therefore, 2 2 41 has the same last digit as 2 4 , which is 6.

Finding the Last Digit of the Result So, the last digit of 2 2 41 − 3 is the last digit of 6 − 3 , which is 3.

Final Answer Thus, 4 2 40 − 3 ends in 3.


Examples
Understanding the last digit of large powers can be useful in cryptography, where certain operations depend on the properties of large numbers. For example, in some encryption algorithms, the last digit of a number might be used to determine a key or to verify the integrity of data. Knowing that 4 2 40 − 3 ends in 3 allows one to quickly assess certain properties without needing to compute the entire value, which can be computationally expensive.

Answered by GinnyAnswer | 2025-07-04