abc is any 3 digit number. Prove algebraically that abc is divisible by 9, if a+b+c is divisible by 9. Is the number 684 divisible by 9? (6+8+4=18) yes it is, so how do you prove it algebraically.

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abc is a any 3-digit number

abc can be written as (100a+10b+c) because a is in the hundreds place, b is in the tens place, and c is in the ones place

for example 684=(600)+(80)+4)

proof: abc=(100a+10b+c)

=(99a+a)+(9b+b)+(c)

=(99a+9b)+(a+b+c)

if two numbers are both divisible by 9 then their sum is divisible by 9

(99a+9b) is divisible by 9 because it can be written as 9(11a+b)

therefore if (a+b+c) is divisible by 9 then (99a+9b)+(a+b+c) is divisible by 9 and we know

that (99a+9b)+(a+b+c)=100a+10b+c=abc, a 3-digit number

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