Question

Докажите что число 2^26+4^11-8^7 делится на 44

Answer 1

2^26 = 67108864
4^11 = 4194304
8^7 = 2097152 Делится: 1572864

Answer 2

To prove that the number $2^{26} + 4^{11} - 8^7$ is divisible by 44, we need to show that it leaves no remainder when divided by 44. In other words, we need to show that $(2^{26} + 4^{11} - 8^7) \mod 44 = 0$.

Let's start by simplifying the expression:

$2^{26} = (2^2)^{13} = 4^{13}$

$4^{11}$ remains unchanged.

$8^7 = (2^3)^7 = 2^{3 \times 7} = 2^{21}$

Now the expression becomes:

$4^{13} + 4^{11} - 2^{21}$

Now, let's consider each term separately with respect to $44$:

1. $4^{13}$ is divisible by 44 because $4^{13} = (2^2)^{13} = 2^{26}$. Since 26 is greater than or equal to 2, $2^{26}$ is divisible by $2^2$, which is 4. Thus, $4^{13}$ is divisible by 4. Similarly, since 26 is greater than or equal to 1, $2^{26}$ is divisible by $2^1$, which is 2. Thus, $4^{13}$ is also divisible by 2. Since $4^{13}$ is divisible by both 4 and 2, it is divisible by 44.

2. $4^{11}$ is divisible by 44 for the same reasons as $4^{13}$. It is divisible by 4 and 2.

3. $2^{21}$ is divisible by 44 because $2^{21} = (2^2)^{10} \times 2 = 4^{10} \times 2$. Again, $4^{10}$ is divisible by both 4 and 2, and when we multiply it by 2, the result is still divisible by 2. Therefore, $2^{21}$ is divisible by 44.

Now, since all three terms $(4^{13} + 4^{11} - 2^{21})$ are divisible by 44, their sum is also divisible by 44. Hence, $2^{26} + 4^{11} - 8^7$ is divisible by 44.