Proof by Induction Prove that for all numbers divisible by 3, the number produced by the sum of their digits is divisible by three (i.e. 261 is divisible by 3, and 2+6+1=9 which is also divisible by 3.
Solution
Suppose n is a positve number whose digit representation is ak...a 3 a 2 a 1 a 0 for example suppose the number is 9876 then a 3 = 9, a 2 = 8, a 1 = 7, a 0 = 6
Now converting the numbers into sum of unit, tens and hundreds place
n = a 0 *1 + a 1 *10 + ...a 0 *10^ k .
The digit sum is s = a 0 + a 1 + ... + a k .
Now we will Compute n - s:
n - s = (a 0 *1 + a 1 *10 + ...a 0 *10^ k .) - (a 0 + a 1 + ... + a k . )
This equals:
a1*9 + a2*99 + ... + ak*(99...9) (where the last term has k nines).
From here you can easily factor out a 3:
3[a1*3 + a2*33 + ... + ak*(33...3)]
This proves that a number less its digit sum is always a multiple of 3 (and 9 for that matter):
Thus n - s is divisible by 3.
Now select a number n such that sum of digits of the number is a multiple of three. And if we add s to n-s then we will get n
If we have two multiples of three like 6 and 9 then also the sum will be divisible by three.
n and s are both multiple of 3 then there difference will be the multiple of three as well
P(n) : a 0 + a 1 + …………… + a n = mod 3
Step 1 for n= 1
1 mod 3 will be zero H.P
Step 2: For n = k, we will assume that the statement is true
a 0 + a 1 + …………… + a k = mod 3
Step 3: For n = k+1 we will prove
a 0 + a 1 + …………… + a k + a k+1 = mod 3
Taking LHS
a 0 + a 1 + …………… + a k = mod 3 (from step 2)
now for a k+1 the next digit will be included in because for two numbers n and s if n and s are divisible by three
n + q(divisible by three) = s (s also divisible by three from above discussion
Hence the LHS also equals to RHS Hence proved
.
Proof by Induction Prove that for all numbers divisible by 3- the numb.docx
1. Proof by Induction Prove that for all numbers divisible by 3, the number produced by the sum of
their digits is divisible by three (i.e. 261 is divisible by 3, and 2+6+1=9 which is also divisible by
3.
Solution
Suppose n is a positve number whose digit representation is ak...a 3 a 2 a 1 a 0 for example suppose
the number is 9876 then a 3 = 9, a 2 = 8, a 1 = 7, a 0 = 6
Now converting the numbers into sum of unit, tens and hundreds place
n = a 0 *1 + a 1 *10 + ...a 0 *10^ k
.
The digit sum is s = a 0 + a 1 + ... + a k .
Now we will Compute n - s:
n - s = (a 0 *1 + a 1 *10 + ...a 0 *10^ k
.) - (a 0 + a 1 + ... + a k . )
This equals:
a1*9 + a2*99 + ... + ak*(99...9) (where the last term has k nines).
From here you can easily factor out a 3:
3[a1*3 + a2*33 + ... + ak*(33...3)]
This proves that a number less its digit sum is always a multiple of 3 (and 9 for that matter):
Thus n - s is divisible by 3.
Now select a number n such that sum of digits of the number is a multiple of three. And if we
add s to n-s then we will get n
If we have two multiples of three like 6 and 9 then also the sum will be divisible by three.
n and s are both multiple of 3 then there difference will be the multiple of three as well
2. P(n) : a 0 + a 1 + …………… + a n = mod 3
Step 1 for n= 1
1 mod 3 will be zero H.P
Step 2: For n = k, we will assume that the statement is true
a 0 + a 1 + …………… + a k = mod 3
Step 3: For n = k+1 we will prove
a 0 + a 1 + …………… + a k + a k+1 = mod 3
Taking LHS
a 0 + a 1 + …………… + a k = mod 3 (from step 2)
now for a k+1 the next digit will be included in because for two numbers n and s if n and s are
divisible by three
n + q(divisible by three) = s (s also divisible by three from above discussion
Hence the LHS also equals to RHS Hence proved
![Proof by Induction Prove that for all numbers divisible by 3, the number produced by the sum of
their digits is divisible by three (i.e. 261 is divisible by 3, and 2+6+1=9 which is also divisible by
3.
Solution
Suppose n is a positve number whose digit representation is ak...a 3 a 2 a 1 a 0 for example suppose
the number is 9876 then a 3 = 9, a 2 = 8, a 1 = 7, a 0 = 6
Now converting the numbers into sum of unit, tens and hundreds place
n = a 0 *1 + a 1 *10 + ...a 0 *10^ k
.
The digit sum is s = a 0 + a 1 + ... + a k .
Now we will Compute n - s:
n - s = (a 0 *1 + a 1 *10 + ...a 0 *10^ k
.) - (a 0 + a 1 + ... + a k . )
This equals:
a1*9 + a2*99 + ... + ak*(99...9) (where the last term has k nines).
From here you can easily factor out a 3:
3[a1*3 + a2*33 + ... + ak*(33...3)]
This proves that a number less its digit sum is always a multiple of 3 (and 9 for that matter):
Thus n - s is divisible by 3.
Now select a number n such that sum of digits of the number is a multiple of three. And if we
add s to n-s then we will get n
If we have two multiples of three like 6 and 9 then also the sum will be divisible by three.
n and s are both multiple of 3 then there difference will be the multiple of three as well](https://image.slidesharecdn.com/proofbyinductionprovethatforallnumbersdivisibleby3-thenumb-230208182318-89c340b5/85/Proof-by-Induction-Prove-that-for-all-numbers-divisible-by-3-the-numb-docx-1-320.jpg)
