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Question Number 212342 by hardmath last updated on 10/Oct/24

$$\mathrm{The}\mathrm{number}\overline{\mathrm{abc}}\mathrm{is}\mathrm{divisible}\mathrm{by}37.$$$$\mathrm{Prove}\mathrm{that}\overline{\mathrm{bca}}+\overline{\mathrm{cab}}\mathrm{is}\mathrm{divisible}\mathrm{by}37.$$

Answered by A5T last updated on 10/Oct/24

$$100a+10b+c\equiv -11a-27b+c=-(11a+27b-c)$$$$\equiv 0\left(mod37\right)$$$$100b+10c+a+100c+10a+b=110c+101b+11a$$$$\equiv 11a+27b-c\equiv \left(0mod37\right)$$

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