/sci/ - Science & Math » Thread #11401914

15KiB, 563x273, equation.png

View Same Google iqdb SauceNAO

Not a homework, just curious

Anonymous Thu 20 Feb 13:43:55 2020 No.11401914 View Reply Original Report

Quoted By: >>11402059 >>11402105 >>11402141 >>11402253 >>11402274 >>11402571 >>11402718

Greetings,

Given the equation in picture, how to find solutions where x,y,z are all whole numbers and z is not 0. Or if no such solution exist how to prove it?

Anonymous

Anonymous Thu 20 Feb 2020 15:04:29 No.11402047 Report

Quoted By: >>11402103

So the sum always gives a number like 111111 with the number of digits equal to z. This always ends in a one.

It ends in a one, so we need 3^x and 7^y to both give numbers ending in a one.
For the 3 times table the product ends in a one when the number you're multiplying by ends in a seven. Like 3*7 = 21, 3*27 = 81.

So we want 3^x to be of the form 3*(3^x-1) where 3^x-1 is a number ending in a 7. We can do the same analysis on 3^x-1 to find that 3^x-2 must end in a 9, and then 3^x-3 must end in a 3, which can just be a three. So 3*3*3*3 ends in a one and we know that 3^x must be of the form 3^4a to give a result ending in a 1.

We can do a similar thing with 7 (or just try putting 7^y in your calc) until we find that we also need 7^y = 7^4b.
Then I would just try a bunch of combinations for 'a' and 'b' until one of them is 111...111.

Anonymous

Anonymous Thu 20 Feb 2020 15:08:04 No.11402059 Report

Quoted By: >>11402078

>>11401914
if x =/= 0, then z mod 3 is necessarily 2

Anonymous

Anonymous Thu 20 Feb 2020 15:15:44 No.11402078 Report

Quoted By: >>11402091

>>11402059
Wrong.

Anonymous

Anonymous Thu 20 Feb 2020 15:21:34 No.11402091 Report

Quoted By: >>11402098

>>11402078
Why?

Anonymous

Anonymous Thu 20 Feb 2020 15:25:29 No.11402098 Report

Quoted By: >>11402104 >>11402111

>>11402091
Different guy here, why would z mod3 = 2?

Anonymous

Anonymous Thu 20 Feb 2020 15:26:54 No.11402103 Report

Quoted By: >>11402117 >>11402118

>>11402047
>It ends in a one, so we need 3^x and 7^y to both give numbers ending in a one.
>For the 3 times table the product ends in a one when the number you're multiplying by ends in a seven. Like 3*7 = 21, 3*27 = 81.
LOL, you don't see how you immediately contradicted yourself? If 3*7 ends in 1 then you don't need 3^x and 7^y to both give numbers ending in a one.

Anonymous

Anonymous Thu 20 Feb 2020 15:26:56 No.11402104 Report

Quoted By:

>>11402098
If you consider the whole equation mod 3 and x > 0 then you get

z + 1 = 0 (mod 3)
or equivalent
z = 2 (mod 3)

Anonymous

Anonymous Thu 20 Feb 2020 15:27:50 No.11402105 Report

Quoted By:

>>11401914

We can rewrite the LHS as (10^(z+1) - 1) / 9

so the equation becomes 10^(z+1) - 1= 3^(x+2) 7^y

might help?

Anonymous

Anonymous Thu 20 Feb 2020 15:29:43 No.11402111 Report

Quoted By:

>>11402098
Because the right side is divisible by 3, so the left side has to be too. And that's the case iff the sum of "1"s is divisible by 3 ->. the number of digits is a multiple of 3 - i.e. when z=2 (111), z=5, ...

Anonymous

Anonymous Thu 20 Feb 2020 15:30:08 No.11402113 Report

Quoted By:

But of course z = 2 (mod 3)

the LHS is divisible by 3 only when the digit sum is divisible by 3.

E.g. 111 and 111111 are divisible by 3.

Anonymous

Anonymous Thu 20 Feb 2020 15:30:39 No.11402117 Report

Quoted By: >>11402148

>>11402103
You have to be 18 to post here.

Anonymous

Anonymous Thu 20 Feb 2020 15:30:57 No.11402118 Report

Quoted By: >>11402127

>>11402103
3*7^2 = 147?

Anonymous

Anonymous Thu 20 Feb 2020 15:35:21 No.11402127 Report

Quoted By:

>>11402118
3*3*7*7 = 21^2 = 441

Anonymous

Anonymous Thu 20 Feb 2020 15:40:53 No.11402141 Report

Quoted By:

>>11401914
10^(z+1) = 3^(x+2) 7^y + 1

There are no solutions

Anonymous

Anonymous Thu 20 Feb 2020 15:42:51 No.11402148 Report

Quoted By:

>>11402117
Please tell me, does 3^1 end in 1? Does 7^1 end in 1? No, so according to your argument 3*7 cannot end in 1.

Anonymous

Anonymous Thu 20 Feb 2020 16:26:23 No.11402253 Report

Quoted By: >>11402264 >>11402284

>>11401914
Left side will be a number ending in 1 because if 10 to the power of 1, thus the right side must end in 1 as well.
Powers of 3 end to 1, 3, 9, 7, the repeat; powers of 7 end to 1, 7, 9, 3 then repeat.
Compare the two together, neither adding nor substracting (negative powers) will yield a number endig with 1, so the right side will never end with a 1, ergo that equation is impossible.

Anonymous

Anonymous Thu 20 Feb 2020 16:29:54 No.11402264 Report

Quoted By:

>>11402253
Small correction: the very first part of the sum is 10 to the power of 0, which is 1

Anonymous

Anonymous Thu 20 Feb 2020 16:34:29 No.11402274 Report

Quoted By:

>>11401914
This problem is pretty easy.
Multiply both sides by 9.
Add 1 to both sides.
Now, you are asking when a power of 10 mod 3 is 1, and when a power of 10 mod 7 is 1. All powers of 10 are congruent to 1 mod 3, so, that doesn't narrow it down, but mod 7 does help. A power of 10 can only be congruent to 1 mod 7 when the exponent is a multiple of 6. Translating this back to the original problem, we have already eliminated all cases where z is not congruent to 5 mod 6. In the cases where z congruent to 5 mod 6, we can inspect the first element (the case where z = 5). 111111's prime factorization happens to include numbers that aren't 3 and 7, so it won't work for z = 5. In the general case, when z is congruent to 5 mod 6, the left hand side can be factorized into a sum of powers of 10^6, and 111111 itself. Because 111111 is always a factor of the left hand side when z = 5 mod 6, and the previous result that z =/= 5 mod 6 can't work, in all cases, this equation can't have any whole number solutions.

Anonymous

Anonymous Thu 20 Feb 2020 16:36:38 No.11402284 Report

Quoted By: >>11402286

>>11402253
but that's wrong. 3*7 = 21

Anonymous

Anonymous Thu 20 Feb 2020 16:38:02 No.11402286 Report

Quoted By: >>11402335

>>11402284
Dammit, I imagined addition instead of multiplying, my bad

Anonymous

Anonymous Thu 20 Feb 2020 16:56:52 No.11402335 Report

Quoted By:

>>11402286
Since I know jackshit about proofing and whatnot, here's my stab at it
Checking the (x,y) pairs that give a 1 in the end of the number have a seemingly cyclic pattern of ending with 01, 21, 41, 81 and then repeat, I'd conclude that there is only one solution, with z=y=x= 0, if summa works like I think it does, if z has to be >0 then there are no solutions. I really need to develop my math for a proper proof on this.

Anonymous

Anonymous Thu 20 Feb 2020 18:28:30 No.11402571 Report

Quoted By: >>11402584 >>11402588 >>11402747

>>11401914
A simplistic solution:
LHS is of the form:111111...1, and LHS must be divisible by 3, so there must be a multiple of 3 of 1's in LHS. Thus LHS must be sivisible by 111 since it is 111+111000+111000000... Therefore RHS is divisible by 111=3*37. 37 is prime and not a prime divisor of RHS, thus a contradiction.

Anonymous

Anonymous Thu 20 Feb 2020 18:31:25 No.11402581 Report

Quoted By: >>11402620

A brute-force solution:

Powers of 3 (mod 100) go like: 3, 9, 27, 81, 43, 29, 87, 61, 83, 49, 47, 41, 23, 69, 7, 21, 63, 89, 67, 1, 3, ... repeating

Powers of 7 (mod 100) go:

1, 7, 49, 43, 1, ... repeating

If we compare all possible multiplications we observe that no product is equal to 11 (mod 100) and thus there is no solution to the equation (beyond the trivial solution x = y = z = 0).

Anonymous

Anonymous Thu 20 Feb 2020 18:32:32 No.11402584 Report

Quoted By:

>>11402571
Nice.

Anonymous

Anonymous Thu 20 Feb 2020 18:34:11 No.11402588 Report

Quoted By: >>11402608 >>11402658

>>11402571
Although for completeness do you need to show no power of 7's is = 11...11 ?

Anonymous

Anonymous Thu 20 Feb 2020 18:38:49 No.11402608 Report

Quoted By: >>11402658

>>11402588
No, we don't need to show, since we know for sure, that it must be divisible by 111. It might be divisble by 111111, 111111111 and so on, but it doesn't matter, because all of those numbers are multiples of 111.

Anonymous

Anonymous Thu 20 Feb 2020 18:41:06 No.11402620 Report

Quoted By:

>>11402581
Did you seriuoslly check all the products? It should work out, but it not nice

Anonymous

Anonymous Thu 20 Feb 2020 18:52:17 No.11402658 Report

Quoted By:

>>11402588
>>11402608
Sorry, misunderstood your question. Yeah, we need to check the case x=0. Standard check of powers of seven mod 100 should be sufficient here.

Anonymous

Anonymous Thu 20 Feb 2020 19:10:34 No.11402718 Report

Quoted By:

>>11401914
First use long division to divide an arbitrarily long string of 1s by 21:

21|1111... = 528(671957)...

You get an infinite repeating string that doesn't contain 0, meaning no string of 1s is divisible by 21. Thus x or y must be 0.

Then use standard modulo arithmetic to show 10^(z+1)-1 =/= 3^(x+2) or 7^y

Anonymous

Anonymous Thu 20 Feb 2020 19:19:14 No.11402747 Report

Quoted By:

>>11402571
Same can be done for 7:

1111... = 111111+11111000000+...

1111111/7 = 3*11*13*37

Since 11, 13, and 37 don't appear in the RHS, they can't be equal.

Capcode	All Only User Posts Only Moderator Posts Only Admin Posts Only Developer Posts
Show Posts	All Only With Images Only Without Images
Deleted Posts	All Only Deleted Posts Only Non-Deleted Posts
Ghost Posts	All Only Ghost Posts Only Non-Ghost Posts
Post Type	All Only Sticky Threads Only Opening Posts Only Reply Posts
Results	All Grouped By Threads
Order	Latest Posts First Oldest Posts First

Your latest searches