Algebra proof
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Dr.Peterson
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Please tell us the context of this problem. It doesn't sound like beginning algebra; do you know anything about modular arithmetic? More generally, what course is this for, and what topics have you studied that might be relevant?
Dr.Peterson
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What competition is this? Can you tell us any more? (And is it the actual contest, or a practice problem?)
The answer will involve some knowledge of divisibility. What have you learned about that topic?
The answer will involve some knowledge of divisibility. What have you learned about that topic?
Its from OMJ (https://omj.edu.pl/uploads/attachments/1etap20.pdf) , but it think I know the answer. The last number of 2^1002 have to be 4 and a*b*c*d=5^1002 which means all 4 numbers (a,b,c,d) are 5 rised to power (5^1=5 5^2=25 5^3=125 5^4=625 e.t.c) so they ends with 5 and the
outcome also ends with 5. It means there are no 4 positive integers that have a sum of 2^1002 and a product of 5^1002 becuse 4≠5. Am I right?
outcome also ends with 5. It means there are no 4 positive integers that have a sum of 2^1002 and a product of 5^1002 becuse 4≠5. Am I right?
Dr.Peterson
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I think you're close; but what if one or more of the numbers is 1? And the sum of four numbers that end in 5 is 0, not 5.
Also, how do you know that the last digit of 2^1002 is 4?
Also, how do you know that the last digit of 2^1002 is 4?
Yes all four numbers must be odd.
Dr. Peterson asked how you know that the units digit in a certain power of 2 is 4. Here is a way to prove it using weak mathematical induction, a technique you probably do not know if you are in primary school.
What we do is that we first prove a proposition is true for a single integer (often 1). We then know that the proposition is true for at least one integer. Take any integer for which the proposition is true (I usually call it k). Now prove it is true for k + 1. If so, it is true for the first number we showed, then the next, then the next, and so on forever and ever.
[MATH]2^1 = 2, \ 2^2 = 4,\ 2^3 = 8,\ 2^4 = 16, \ 2^5 = 32. [/MATH]
So we know that
[MATH]n = 1 \implies 2^{4n - 2} = 2^2 = 10 * 0 + 4 = 10i_n + 4 \text { and } i_n \in \mathbb N_{\ge 0}.[/MATH]
[MATH]k \text { is any integer} \ge 1 \text { and such that }[/MATH]
[MATH]2^{(4k-2)} = 10i_k + 4 \text { and } i_k \in \mathbb N_{\ge 0}.[/MATH]
[MATH]2^{\{4(k+1)- 2\}} = 2^{(4k+4-2)} = 2^{\{(4k-2)+4\}} =[/MATH]
[MATH]2^{(4k-2)} * 2^4 = (10i_k + 4) * 16 = 16(10i_k) + 64 = [/MATH]
[MATH]10(16i_k) + 10(6) + 4 = 10(6 + 16i_k) + 4.[/MATH]
[MATH]i_k \text { is a non-negative integer} \implies 6 + 16i_k \text { is too.}[/MATH]
[MATH]\text {Let } i_{k+1} = 6 + 16i_k \implies[/MATH]
[MATH]2^{\{4(k+1)-2)} = 10i_{(k+1)} + 4 \text { and } i_{(k+1)} \in \mathbb N_{\ge 0}.[/MATH]
[MATH]\therefore n \in \mathbb N_{> 0} \implies 2^{4n - 2} = 10i_n + 4 \text { and } i_n \in \mathbb N_{\ge 0}.[/MATH]
All that remains to do is to show that
[MATH]1002 = 1004 - 2 = 4 * 251 - 2.[/MATH]
And clearly 251 is a non-negative integer.
I doubt a primary school student is expected to do that proof. (Or you could use modular arithmetic if you know that).
Can all four numbers equal 1?
If just three of the numbers are 1, what is the unit digit of the sum of all four numbers?
Dr. Peterson asked how you know that the units digit in a certain power of 2 is 4. Here is a way to prove it using weak mathematical induction, a technique you probably do not know if you are in primary school.
What we do is that we first prove a proposition is true for a single integer (often 1). We then know that the proposition is true for at least one integer. Take any integer for which the proposition is true (I usually call it k). Now prove it is true for k + 1. If so, it is true for the first number we showed, then the next, then the next, and so on forever and ever.
[MATH]2^1 = 2, \ 2^2 = 4,\ 2^3 = 8,\ 2^4 = 16, \ 2^5 = 32. [/MATH]
So we know that
[MATH]n = 1 \implies 2^{4n - 2} = 2^2 = 10 * 0 + 4 = 10i_n + 4 \text { and } i_n \in \mathbb N_{\ge 0}.[/MATH]
[MATH]k \text { is any integer} \ge 1 \text { and such that }[/MATH]
[MATH]2^{(4k-2)} = 10i_k + 4 \text { and } i_k \in \mathbb N_{\ge 0}.[/MATH]
[MATH]2^{\{4(k+1)- 2\}} = 2^{(4k+4-2)} = 2^{\{(4k-2)+4\}} =[/MATH]
[MATH]2^{(4k-2)} * 2^4 = (10i_k + 4) * 16 = 16(10i_k) + 64 = [/MATH]
[MATH]10(16i_k) + 10(6) + 4 = 10(6 + 16i_k) + 4.[/MATH]
[MATH]i_k \text { is a non-negative integer} \implies 6 + 16i_k \text { is too.}[/MATH]
[MATH]\text {Let } i_{k+1} = 6 + 16i_k \implies[/MATH]
[MATH]2^{\{4(k+1)-2)} = 10i_{(k+1)} + 4 \text { and } i_{(k+1)} \in \mathbb N_{\ge 0}.[/MATH]
[MATH]\therefore n \in \mathbb N_{> 0} \implies 2^{4n - 2} = 10i_n + 4 \text { and } i_n \in \mathbb N_{\ge 0}.[/MATH]
All that remains to do is to show that
[MATH]1002 = 1004 - 2 = 4 * 251 - 2.[/MATH]
And clearly 251 is a non-negative integer.
I doubt a primary school student is expected to do that proof. (Or you could use modular arithmetic if you know that).
Can all four numbers equal 1?
If just three of the numbers are 1, what is the unit digit of the sum of all four numbers?
Dr.Peterson
Elite Member
- Joined
- Nov 12, 2017
- Messages
- 16,093
The heading of the document says
That suggests that we shouldn't be giving specific help for a few days. It may suggest more than that.
XVI Junior Mathematics Olympiad (2020/21) Competition tasks of the first level - correspondence part (September 1 - October 12, 2020)
That suggests that we shouldn't be giving specific help for a few days. It may suggest more than that.
Thanks a lot, for help, I understand maybe I shoudn't post it here. Unlike many others people I did all exercises exept that one 100% myself. I really don't have anyone who would help me (parents, tutors, teachers e.t.c). The rules also don't say help from other people is forbidden. I don't just copy and paste this exercise so please don't get me wrong I just wanted for someone to guide me a bit, not make it for me.
