Digital index
Given 2 three digit numbers a and b and a four digit number c . If the sum of the digits of the number a + b , b + c , c + a are all equal to 3, find the largest possible sum of the digits of the number a + b + c .
The answer is 18.
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1 solution
Let a = 4 5 6 , b = 5 4 6 , and c = 1 5 5 4 . Then a + b = 1 0 0 2 , a + c = 2 0 1 0 , and b + c = 2 1 0 0 ; the sum of the digits of each number is 3. Also, a + b + c = 2 5 5 6 , and the sum of the digits of 2556 is 2 + 5 + 5 + 6 = 1 8 .
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Hey!You are very much right till the very last step!But, you have to perform the sum,repeatedly(that is,iteratively). YOUR final answer is 18.But 1+8=9.(which is the correct answer.
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The problem does not say anything about adding up the digits repeatedly. If this is the intent of the problem, then it should be edited to reflect this.
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@Jon Haussmann – From my experience,I have found that in questions like this,we have to preform the sum repeatedly.It is understood for those who are familiar with these type of problems,but yes, it should be specified in the problem.
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@Anandmay Patel – No, performing the sum repeatedly is known as the "digit root" and not the "digit sum".
How did you find these numbers?
I just concluded that the answer must be 9 or 18 or 27 and there were 3 tries, so...
The sum of the digits of the numbers a+b,b+c and c+a are all equal to 3.This means that they all are divisible by 3. SO (a+b)+(b+c)+(c+a) is divisible by 3. That is,2(a+b+c) is divisible by 3. So a+b+c is divisible by 3. If we perform iterative processes of adding up the digits of a number,it will always end up in a single digit number.[FOR EXAMPLE, Digital sum of 2344567 performed repeatedly is=2+3+4+4+5+6+7=31.AND 3+1=4(a single digit number). But in our case,the number(a+b+c)is divisible by 3.So if we apply an iterative digital sum to a number divisible by 3,it will end in a single digit number,divisible by 3(by the converse of divisibility laws for 3). 9 can be the maximum such number.