51 times of 51 minus one is divided by 7

51 = 49 + 2, so after the expansion of 51 to the 51st power of 2, all the other items contain 49, and the items containing 49 can be divisible by 7. In this way, the problem is transformed into proving that: (1) the 51st power of 2 can be divisible by 7. The 51st power of 2 = 2 × 2 to the 50th power = 2 × 4 to the 25th power = 8 × 4 to the 24th power = (7 + 1) × 4 to the 24th power

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