If sum of digits div. by 3, then number itself div. by 3

[karmsy]

precalculus: number series (I think): FYI, I've been using two precalculus textbooks, one by Sullivan, the other by Lothar/Redlin/Watson.

This question doesn't come from either textbook, but from a study guide somebody put together on a private basis, with solutions. I didn't find the solution given for this problem particularly helpful. I'm hoping users of this forum can do better:

"If the sum of the digits of a number is divisible by 3, then the number itself is divisible by 3."

What would you do with something like that?

 

[tkhunny]

Re: precalculus: number series (I think)

It is a well-known test for divisibility. Sometimes you just want to know if things are divisible.

If you have 51 apples and you wish to divide them evenly amongst three (3) people, will you be able to do it without cutting an apple?

 

[mmm4444bot]

Re: precalculus: number series (I think)

... What would you do with something like that?

Use it to quickly see that fractions containing multiples of 3 can be reduced.

For example, 33/1242.

If I would like to know whether or not 33/1242 reduces, then I have to wonder if either 3 or 11 goes into 1242 evenly. The divisibility rule lets me settle this question mentally and quickly.

To use TK's numerals, the divisibility rule allows me to "see" that 51/3 can be reduced to a whole number.

You wrote, "number series".

What exactly are you looking at? Are you looking at a proof of the divisibility rule?

I ask because you noted that you did not find the problem solution very helpful.

 

[Deleted member 4993]

Re: precalculus: number series (I think)

What would you do with something like that?

That will tell you quickly - whether a number is not a prime number.

For example this rule tells me 51 (5+1 = 6) is not a prime number - neither is 57.

However, for 53 it can only tell it is not divisible by 3.

There some interesting rules to to find divisibility by 6 [unit number must be even (divisible by 2) and digits add up to be divisible by 3), divisibility by 9 (digits add up to be divisible by 9), divisibility by 7 and divisibility by 11.

These rules become important in locating prime numbers - that is used in cryptography. You may want read - the Code Book by Simon Singh.

 

[karmsy]

Re: precalculus: number series (I think)

Whoops. I didn't ask my question very well. No wonder the answers weren't of very good quality (no offense

I was supposed to PROVE the proposition that "If the sum of the digits of a number is divisible by 3, then the number itself is divisible by 3."

Could somebody help me here? Sorry for the confusion.

 

[Deleted member 4993]

Re: precalculus: number series (I think)

Whoops. I didn't ask my question very well. No wonder the answers weren't of very good quality (no offense

I was supposed to PROVE the proposition that "If the sum of the digits of a number is divisible by 3, then the number itself is divisible by 3."

Could somebody help me here? Sorry for the confusion.

What level math are you doing?

Please share with us your work/thoughts - so that we know where to begin to help you.

 

[mmm4444bot]

Re: precalculus: number series (I think)

... This question ... [comes] from a study guide ... with solutions ...

... I was supposed to PROVE the proposition ...

Yes, I agree with Subhotosh. I would like to know whether or not you understand the notation of the answer that you see in the study guide.

Maybe you understand it somewhat, and you need clarification on one of the steps?

Or maybe you don't understand it at all?

It's hard to know where to start.

~ Mark

 

[karmsy]

Re: precalculus: number series (I think)

"If the sum of the digits of a number is divisible by 3, then the number itself is divisible by 3."

Here we go with the proof that was provided, that didn't help me:

"Let the number N be represented as: a(subscript n)a(sub n-2)...a(sub 3)a(sub 2)a(sub 1), where the a(sub i) represents the actual digits of N, so that a(sub n)a(sub n-2)...a(sub 3)a(sub 2)a(sub 1) is actually a string of digits constituting N, like N=7498.

Then the value of N=10(superscript n-1)a(subscript n) + 10(superscript n-2)a(subscript n-1)...just as 7498=7x1000 + 4x100 + 9x10 +8 (1)

Since the sum of the digits of N: a(sub n) + a(sub n-2)...a(sub 3) + a(sub 2) + a(sub 1), is divisible by 3, that makes the sum a multiple of 3. That is,
a(sub n) + a(sub n-2)...a(sub 3) + a(sub 2) + a(sub 1)= 3K, where K is an arbitrary number. (2)

a(subscript 1) = 3K - a(sub n) - a(sub n-2)...a(sub 3) - a(sub 2) - a(sub 1) (3)

substituting 3 into 2 and simplifying:

N=(10(superscript n-1)-1)a(subscript n) + (10(superscript n-2)-1)a(subscript n-1)...99a(subscript 3) + 9a(subscript 2) (4)

Now, 10(superscript n-1) - 1 has (n-1)9s, 10(superscript n-2) -1 has (n-2)9s, and so on, so that each term in (4) is divisible by 3.

Therefore, N itself is divisible by 3."

'Bout the only thing that I understand in this whole proof is the example of 7498, i.e., we weight each digit in a number according to its place value. The rest is over my head. And I apologize for not reproducing many specialized keyboard symbols that appeared in the original. I hope my text substitutions were clear.

 

[Deleted member 4993]

Re: precalculus: number series (I think)

I would have proved almost the same way the book has done.

Without knowing your level of math (or what you-should-have-already-known) I cannot do a better job. Your best bet would be to seek the assistance of a face-to-face tutor.

 

[karmsy]

Re: precalculus: number series (I think)

I have a solid working knowledge of high-school math: beginning and intermediate algebra, geometry, and trigonometry. I also have a basic understanding, though less familiarity, with calculus. I have worked a reasonable amount recently with number series: arithmetic and geometric progressions, the binomial theorem, and mathematical induction. I understand the proof given above involved some kind of mathematical induction, but the logic just doesn't "flow" for me.

 

[Deleted member 4993]

Re: precalculus: number series (I think)

"If the sum of the digits of a number is divisible by 3, then the number itself is divisible by 3."

Here we go with the proof that was provided, that didn't help me:

"Let the number N be represented as: a(subscript n)a(sub n-2)...a(sub 3)a(sub 2)a(sub 1), where the a(sub i) represents the actual digits of N,

N --- a[sub:3p9eykam]n[/sub:3p9eykam]a[sub:3p9eykam]n-1[/sub:3p9eykam]a[sub:3p9eykam]n-2[/sub:3p9eykam]...a[sub:3p9eykam]3[/sub:3p9eykam]a[sub:3p9eykam]n[/sub:3p9eykam]2[sub:3p9eykam]1[/sub:3p9eykam]

so that a(sub n)a(sub n-2)...a(sub 3)a(sub 2)a(sub 1) is actually a string of digits constituting N, like N=7497.

Then the value of N=10(superscript n-1)a(subscript n) + 10(superscript n-2)a(subscript n-1)...just as 7497=7x1000 + 4x100 + 9x10 +7 (1)<<< for 7497 we have n = 4

and

7497 = 10[sup:3p9eykam]4-1[/sup:3p9eykam] * 7 + 10[sup:3p9eykam]3-1[/sup:3p9eykam] * 4 + 10[sup:3p9eykam]2-1[/sup:3p9eykam] * 9 + 10[sup:3p9eykam]1-1[/sup:3p9eykam] * 7

and

a[sub:3p9eykam]4[/sub:3p9eykam] = 7

a[sub:3p9eykam]3[/sub:3p9eykam] = 4

a[sub:3p9eykam]2[/sub:3p9eykam] = 9

a[sub:3p9eykam]1[/sub:3p9eykam] = 7

Since the sum of the digits of N: a(sub n) + a(sub n-2)...a(sub 3) + a(sub 2) + a(sub 1), is divisible by 3, that makes the sum a multiple of 3. That is,

a(sub n) + a(sub n-2)...a(sub 3) + a(sub 2) + a(sub 1)= 3K,

where K is an arbitrary number. (2)<<<< in this case 7+4+9+7 = 27 >>> so K = 9

N = 10[sup:3p9eykam]n-1[/sup:3p9eykam] * a[sub:3p9eykam]n[/sub:3p9eykam] + 10[sup:3p9eykam]n-2[/sup:3p9eykam] * a[sub:3p9eykam]n-1[/sub:3p9eykam] +......+ 10[sup:3p9eykam]1[/sup:3p9eykam] * a[sub:3p9eykam]2[/sub:3p9eykam] + 10[sup:3p9eykam]0[/sup:3p9eykam] * a[sub:3p9eykam]1[/sub:3p9eykam]

N - (a[sub:3p9eykam]n[/sub:3p9eykam] + a[sub:3p9eykam]n-1[/sub:3p9eykam]... + a[sub:3p9eykam]2[/sub:3p9eykam] + a[sub:3p9eykam]2[/sub:3p9eykam]) = 10[sup:3p9eykam]n-1[/sup:3p9eykam] * a[sub:3p9eykam]n[/sub:3p9eykam] + 10[sup:3p9eykam]n-2[/sup:3p9eykam] * a[sub:3p9eykam]n-1[/sub:3p9eykam] +......+ 10[sup:3p9eykam]1[/sup:3p9eykam] * a[sub:3p9eykam]2[/sub:3p9eykam] + 10[sup:3p9eykam]0[/sup:3p9eykam] * a[sub:3p9eykam]1[/sub:3p9eykam] - (a[sub:3p9eykam]n[/sub:3p9eykam] + a[sub:3p9eykam]n-1[/sub:3p9eykam]... + a[sub:3p9eykam]2[/sub:3p9eykam] + a[sub:3p9eykam]1[/sub:3p9eykam])

N - 3K = (10[sup:3p9eykam]n-1[/sup:3p9eykam]-1) * a[sub:3p9eykam]n[/sub:3p9eykam] + (10[sup:3p9eykam]n-2[/sup:3p9eykam] -1)* a[sub:3p9eykam]n-1[/sub:3p9eykam] +......+ (10[sup:3p9eykam]1[/sup:3p9eykam] -1)* a[sub:3p9eykam]2[/sub:3p9eykam] + (10[sup:3p9eykam]0[/sup:3p9eykam] -1)* a[sub:3p9eykam]1[/sub:3p9eykam]

N - 3K = (10[sup:3p9eykam]n-1[/sup:3p9eykam]-1) * a[sub:3p9eykam]n[/sub:3p9eykam] + (10[sup:3p9eykam]n-2[/sup:3p9eykam] -1)* a[sub:3p9eykam]n-1[/sub:3p9eykam] +......+ (9)* a[sub:3p9eykam]2[/sub:3p9eykam] + (0)* a[sub:3p9eykam]1[/sub:3p9eykam]

Every term on the RHS is divisible by 3 (actually by 9) - let us write RHS = 3M

Then

N - 3K = 3M

N = 3(K+M)

Q.E.D

If you don't understand the logic above - tell me exactly where I loose you first....


a(subscript 1) = 3K - a(sub n) - a(sub n-2)...a(sub 3) - a(sub 2) - a(sub 1) (3)

substituting 3 into 2 and simplifying:

N=(10(superscript n-1)-1)a(subscript n) + (10(superscript n-2)-1)a(subscript n-1)...99a(subscript 3) + 9a(subscript 2) (4)

Now, 10(superscript n-1) - 1 has (n-1)9s, 10(superscript n-2) -1 has (n-2)9s, and so on, so that each term in (4) is divisible by 3.

Therefore, N itself is divisible by 3."

'Bout the only thing that I understand in this whole proof is the example of 7498, i.e., we weight each digit in a number according to its place value. The rest is over my head. And I apologize for not reproducing many specialized keyboard symbols that appeared in the original. I hope my text substitutions were clear.

 

[mmm4444bot]

Re: precalculus: number series (I think)

... the logic just doesn't "flow" for me.

Hi Karmsy:

I hope Subhotosh's example helps you. Earlier, you wrote that each digit is "weighted" in the proof you posted. I'm not sure what you're thinking, but none of the digits is any more significant than another. I would guess that the notation is throwing you because the strategy behind the "logic" is actually very basic.

The proof comes from the fact that after subtracting 1 from any whole power of 10, the result is always divisible by 3. (Actually, as Subhotosh noted, it's divisible by 9 as well.) Or, looked at another way, if you divide any whole power of 10 by 3, you will always get a remainder of 1.

Also, we need to realize that if you add two quantities -- both of which are divisible by 3 -- then the sum must also be divisible by 3 because multiplication is nothing more than repeated addition.

The strategy of this proof is to rewrite the power series representing the number N as a sum of two quantities: (A) the sum of the number's digits and (B) whatever's left over (with a 9 factored out). Please see my example below.

Do you find the notation awkward to read? I do, sometimes. The notation must show n-1 and so forth in both the subscripts and exponents in order to deal with any given number of digits. When I experience difficulty "seeing" exactly what notation represents, I find that it helps me to write out a specific case with concrete subscripts and exponents. I'll do that now for the four-digit case, and maybe it will help you to look at this.

FOUR-DIGIT CASE

N is represented by a[sub:1u9j2cah]4[/sub:1u9j2cah]a[sub:1u9j2cah]3[/sub:1u9j2cah]a[sub:1u9j2cah]2[/sub:1u9j2cah]a[sub:1u9j2cah]1[/sub:1u9j2cah]

Therefore, the sum of the digits of N = a[sub:1u9j2cah]4[/sub:1u9j2cah] + a[sub:1u9j2cah]3[/sub:1u9j2cah] + a[sub:1u9j2cah]2[/sub:1u9j2cah] + a[sub:1u9j2cah]1[/sub:1u9j2cah]

The value of N can be written as a power series.

N = 1000 * a[sub:1u9j2cah]4[/sub:1u9j2cah] + 100 * a[sub:1u9j2cah]3[/sub:1u9j2cah] + 10 * a[sub:1u9j2cah]2[/sub:1u9j2cah] + a[sub:1u9j2cah]1[/sub:1u9j2cah]

I now rewrite the powers of 10 using the "ones" that I mentioned above.

N = (1 + 999) * a[sub:1u9j2cah]4[/sub:1u9j2cah] + (1 + 99) * a[sub:1u9j2cah]3[/sub:1u9j2cah] + (1 + 9) * a[sub:1u9j2cah]2[/sub:1u9j2cah] + a[sub:1u9j2cah]1[/sub:1u9j2cah]

Next, I factor and regroup terms in the power series as a sum of two quantities.

N = (a[sub:1u9j2cah]4[/sub:1u9j2cah] + a[sub:1u9j2cah]3[/sub:1u9j2cah] + a[sub:1u9j2cah]2[/sub:1u9j2cah] + a[sub:1u9j2cah]1[/sub:1u9j2cah]) + (999 * a[sub:1u9j2cah]4[/sub:1u9j2cah] + 99 * a[sub:1u9j2cah]3[/sub:1u9j2cah] + 9 * a[sub:1u9j2cah]2[/sub:1u9j2cah])

It's clear that 9 can be factored out of the second quantity.

N = (a[sub:1u9j2cah]4[/sub:1u9j2cah] + a[sub:1u9j2cah]3[/sub:1u9j2cah] + a[sub:1u9j2cah]2[/sub:1u9j2cah] + a[sub:1u9j2cah]1[/sub:1u9j2cah]) + 9 * (99 * a[sub:1u9j2cah]4[/sub:1u9j2cah] + 9 * a[sub:1u9j2cah]3[/sub:1u9j2cah] + a[sub:1u9j2cah]2[/sub:1u9j2cah])

The value of the quantity in red is unimportant; I will replace this number with the symbol K (just like Subhotosh did in his version).

N = (a[sub:1u9j2cah]4[/sub:1u9j2cah] + a[sub:1u9j2cah]3[/sub:1u9j2cah] + a[sub:1u9j2cah]2[/sub:1u9j2cah] + a[sub:1u9j2cah]1[/sub:1u9j2cah]) + 9 * K

I can stop now because 9K is clearly divisible by 3.

Therefore, IF the sum of these digits is divisible by 3, THEN adding another number which is also divisible by 3 (such as 9K) will result in a number N that is divisible by 3.

Please let us know if you still have questions.

Cheers,

~ Mark

PS: This also shows that a number is divisible by 9 if the sum of its digits is divisible by 9.