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Ashish Khandelwal.
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- July 1, 2018 at 9:36 pm #21649
Dr. Ashani DasguptaKeymasterPost your solution here:
Rigorously prove that if a number is divisible by 3 then the sum of its digits is divisible by 3
(check in different bases)
July 1, 2018 at 9:53 pm #21653Pinaki BiswasParticipantSir, here is the proof:
Given: The number is an INTEGER.
To Prove: If a number is divisible by 3, then the SUM of the DIGITS is divisible by 3.(Converse)
Proof: Suppose there exists a number abc with a as HUNDREDS PLACE, b as TENS and c as ONES.
- abc= 100a+10b+c
Suppose a+b+c=3n where n is an INTEGER.
Divide 100a+10b+c by 3.
You will get the REMAINDER as a+b+c.
a+b+c is divisible by 3.
Therefore, 100a+10b+c is DIVISIBLE by 3.
So, abc is DIVISIBLE by 3. (Proved) 🙂
July 3, 2018 at 10:00 pm #21657Shahbaz KhanParticipantUh, I'm not sure how I'm supposed to type math here, so I'll just type in the concept of how I would prove this question.
So, any integer can be written as the sum of various terms of the form a*(10^k) where k is a whole number, a is an integer belonging to [0,9]
Now, we have to note that 10 is congruent to 1 modulo 3. Keeping this in mind, we can assert that in the above terms, the remainder when dividing each by 3 will be the constant, that is, the "a" term, and hence, the entire number is congruent to the sum of the constant terms modulo 3.
Now, the initial number and the sum of the constants is the same modulo 3. That is, when the number is divisible by 3, the sum will be divisible too.
This proves the given statement.
July 6, 2018 at 10:50 pm #21671Writaban SarkarParticipantLet the number \(n=\) \(d_k \cdot 10^k+d_{k-1}\cdot 10^{k-1}+ . . .+d_1\cdot 10^1+d_0\cdot 10^0 \) . If we divide \(n\) by 3 , we get \(d_k+d_{k-1}+ . . .+d_1+d_0\) as remainder. Therefore , for \(n\) to be divisble by 3 the sum of it digits i.e. \(d_k+d_{k-1}+ . . .+d_1+d_0\) must divisible by 3.
Hence proven.
July 7, 2018 at 8:07 pm #21673Ashish KhandelwalParticipantSuppose that you have a four-digit number nn that is written abcdabcd. Then
n=103a+102b+10c+d=(999+1)a+(99+1)b+(9+1)c+d=(999a+99b+9c)+(a+b+c+d)=3(333a+33b+3c)+(a+b+c+d),n=103a+102b+10c+d=(999+1)a+(99+1)b+(9+1)c+d=(999a+99b+9c)+(a+b+c+d)=3(333a+33b+3c)+(a+b+c+d),
so when you divide nn by 33, you’ll get333a+33b+3c+a+b+c+d3.333a+33b+3c+a+b+c+d3.
The remainder is clearly going to come from the division a+b+c+d3a+b+c+d3, since 333a+33b+3c333a+33b+3c is an integer.Now generalize: make a similar argument for any number of digits, not just four. (If you know about congruences and modular arithmetic, you can do it very compactly.)
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