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ABC is a three digit number. It is known that 2A + 3B + C can be divisible by 7. It is proved that this three digit number can also be divisible by 7
This three digit number is 100A + 10B + C
Since 2A + 3B + C can be divided by 7, let 2A + 3B + C = 7n
Then C = 7n-2a-3b
So 100A + 10B + C = 100A + 10B + 7n-2a-3b = 98a + 7b + 7n = 7 (14a + B + n)
Because 7 (14a + B + n) is divisible by 7, 100A + 10B + C is divisible by 7
That is, ABC can be divided by 7
Have a good time
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