Number Theory Puzzler: Proving N-S is a Multiple of 9

[sizzlaw]

Given: N is a four digit number. S is the sum of N's digits.

Prove: N minus S is a multiple of 9.

 

[Goongyae]

Call the four digits of N "abcd"
Then N is (a * 1000 + b * 100 + c * 10 + d)
And N - S = N - a - b - c - d = (a * 999 + b * 99 + c * 9)
And (N - S)/9 is (a * 111 + b * 11 + c), which is an integer.
Hence N-S is divisible by 9.

 

[sizzlaw]

Revised puzzler:

Given: N is a four digit number. S is the sum of N's digits.

Prove: The sum of the digits of (N - S) is divisible by 9.

 

[micromass]

This follows easily from Goongyae's post, since a number is divisible by 9 if and only if the sum of it's digits is divisible by 9.

This is real easy to prove: take x₁...x_n be an n-digit number, then

$$x_110^{n-1}+...+x_n10^0=x_1+...+x_n~(mod~9)$$.

 

[blue_raver22]

I would approach this problem by first understanding the concept of divisibility and its properties. In this case, we are dealing with the divisibility rule for 9, which states that a number is divisible by 9 if the sum of its digits is also divisible by 9.

Now, let's assume that N is a four digit number, represented as ABCD, where A, B, C, and D are the individual digits. This can also be written as N = 1000A + 100B + 10C + D.

Similarly, S can be represented as A + B + C + D.

Now, let's consider N - S:

N - S = (1000A + 100B + 10C + D) - (A + B + C + D)
= (999A + 99B + 9C)

We can see that N - S is a multiple of 9, as it is divisible by 9 with no remainder. This can be further simplified as 9(111A + 11B + C).

Since 111A + 11B + C is an integer, we can say that N - S is a multiple of 9, proving our initial statement.

In conclusion, we have proven that N - S is a multiple of 9, using the properties of divisibility and the representation of N and S in terms of their individual digits. This result can also be extended to any number of digits, as long as the same principles are applied.