Mathematics
For three 3-digit numbers abc, cab and bca, show that abc + cab + bca is divisible by 37.
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Answer
abc + cab + bca
= (100a + 10b + c) + (100c + 10a + b) + (100b + 10c + a)
= 100a + 10b + c + 100c + 10a + b + 100b + 10c + a
= (100a + 10a + a) + (10b + b + 100b) + (c + 100c + 10c)
= 111a + 111b + 111c
= 111(a + b + c)
= 3 x 37 x (a + b + c)
Hence, (abc + cab + bca) is divisible by 37
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