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If one uses the binary exponentiation (exponentiation by squaring) method to compute $a^{55}$, which of the following intermediate powers of $a$ is calculated but NOT used in the final product?

  1. $a^4$
     
  2. $a^8$
     
  3. $a^{16}$
     
  4. $a^{32}$

2 Answers

4 4 votes

Convert the Exponent to Binary

First, we find the binary representation of the exponent, which is 55 .

$$
55_{10}=110111_2
$$


Express as a Sum of Powers of 2

The binary representation $110111_2$ shows us which powers of 2 sum up to 55 :

$$
55=(1 \cdot 32)+(1 \cdot 16)+(0 \cdot 8)+(1 \cdot 4)+(1 \cdot 2)+(1 \cdot 1)
$$


Therefore, the final product to get $a^{55}$ is:

$$
a^{55}=a^{32} \cdot a^{16} \cdot a^4 \cdot a^2 \cdot a^1
$$


Identify Calculated vs. Used Powers
To get the terms needed for the final product, we must calculate a sequence of powers by repeated squaring:

  • $a^1$
     
  • $a^2=a^1 \cdot a^1$
     
  • $a^4=a^2 \cdot a^2$
     
  • $a^8=a^4 \cdot a^4$
     
  • $a^{16}=a^8 \cdot a^8$
     
  • $a^{32}=a^{16} \cdot a^{16}$

As you can see, $a^8$ must be calculated to find $a^{16}$, but because there is a ' 0 ' in the 8 's place of the binary representation of 55 , it is not used in the final multiplication.

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