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Suppose a 3-digit number abc is divisible by 3. Prove thatabc+bca+cab is divisible by 9 |
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Answer» Abc is divisible by 3 => a+b+c = 3k for some integer k. sum of threenumbersabc ,bca ,cab= 100 a+ 10 b+ c + 100 b+ 10c+ a+ 100 c+ 10 a+ b= 100 (a+b+c) + 10 (b+c+a) + (c+a+b) = (a+b+c) * 111 = 3 k * 3 * 37 = 9 * 37 * k Hence, proved. for a no. to be divisible by 3. the sum of digits should be multiple of 3. so. the sum = a+b+c . and in all the three below no. i.e abc , bca and cab the sum of no. will be same.. that is (a+ b+ c) + ( b+c+a) + ( c+a+b) = 3(a+b+c) .and since a+b+c is divisible by 3. so multiple pf a+b+c by 3, will be divisible by 3×3= 9. thank you so much |
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