If a, b, c are three digits of a three-digit number, prove that abc + cab + bca is a multiple of 37.
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abc+bca+cab=(100a+10b+c)+(100b+10c+a)+(100c+10a+b)
abc+bca+cab=(100a+10b+c)+(100b+10c+a)+(100c+10a+b)=111a+111b+111c
abc+bca+cab=(100a+10b+c)+(100b+10c+a)+(100c+10a+b)=111a+111b+111c=111(a+b+c)
abc+bca+cab=(100a+10b+c)+(100b+10c+a)+(100c+10a+b)=111a+111b+111c=111(a+b+c)=3×37(a+b+c)
abc+bca+cab=(100a+10b+c)+(100b+10c+a)+(100c+10a+b)=111a+111b+111c=111(a+b+c)=3×37(a+b+c)Therefore the number abc+bca+cab is divisible by3,37 and (a+b+c) but not by 9.
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