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We know that the first power of 2 = 2, the second power = 4, the third power of 2 = 8,... (please explain it carefully, because I want to give a speech) (1) Can you infer from this the number of the 64 th power of 2? (2) According to the above conclusion, combined with the calculation, please estimate the numbers of (2 + 1) × (2 + 1) × (4 power of 2 + 1). (64 power of 2 + 1)? The second question, please use the square difference formula to solve!
It can be seen that the quintic power of 2 is 326, which is more than 64, and so on. By multiplying by 2, we can see that the number of digits of 6 is the same as that of the first power, the number of digits of the second power is the same as that of the sixth power, and so on. Because there is no carry, every multiplying by two is the same cycle of the fourth power. Then the 64 power is equivalent to the 16 power, which is equivalent to the fourth power
(2) It can be seen from the above that there are four 3 579 digits after adding 1. Similarly, because the product of a digit does not need carry, because the product of 5 and an odd number is always 5, the result of 3 579 is 5 no matter how many times
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