Questions which arose from "A digit reversal"

Last week, I set the following problem - A digit reversal, which confused many people (esp if you read the solutions). A high proportion of people first gave an answer of around 960. Many even claimed that 959 is the correct answer, and gave the following justification:

Since T2S2=(TS)(T+S) T^2 - S^2 = (T-S) (T+S), we can check that the last three digits of T+S T+S are 421 and the last three digits of TS T-S are 779. Hence, the last three digits of T2S2 T^2 - S^2 are obtained from (T+S)×(TS)=421×779=327959 (T+S) \times (T-S) = 421 \times 779 = 327959 . Thus the answer is 959.

What is wrong with the above solution? What was the wrong assumption / misconception that was made?

(Note that I am not asking for your solutions; you can view the solution discussion for different approaches.)


Here are some questions to guide you.
1. What is the remainder when 123 is divided by 1000?
2. What is the remainder when -123 is divided by 1000?
3. What are the last three digits of 123?
4. What are the last three digits of -123?
5. What is 123(mod1000) 123 \pmod{1000} ?
6. What is 123(mod1000) -123 \pmod{1000} ?

Note by Calvin Lin
6 years, 1 month ago

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21 votes

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I'm confused, how does one even get TST - S = 779? If you consider the numbers (mod1000) \pmod {1000} then you get T=321,S=100T = 321, S = 100 thus TS=221T - S = 221 and T+S=421T + S = 421? I'm confused as to how you can even get 779 as an answer (I realize it's 221(mod1000)-221 \pmod {1000} but I still don't know how you would get that). I'm probably missing something obvious.

As for the solution you posted, the problem is that T2S2T^2 - S^2 is negative, thus even though the last 3 digits of it are 959959, 95941(mod1000) -959 \equiv 41 \pmod {1000}.

Michael Tong - 6 years, 1 month ago

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Great, that is one of the strange aspects of this problem.

When trying to determine the last 3 digits of a number, you need to look at several of the last values. I believe that they looked at the last few digits, namely 43219100=4779 4321 - 9100 = - 4779 or 5432199100=44779 54321 - 99100 = -44779 , to 'conclude' that it must be 779.

I'm not commenting if 779 is indeed the last 3 digits, merely on my interpretation of the writeup.

Calvin Lin Staff - 6 years, 1 month ago

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The last 3 digits of T-S are 221, not 779. 321-100=221

Mani Jha - 6 years, 1 month ago

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Can you justify that claim?

What are the last 3 digits of 43219100 4321 - 9100 ? Why is there a sudden change?

Calvin Lin Staff - 6 years, 1 month ago

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Oh, I was wrong.Since T<S, the last digits of TST-S will be equal to the last digits of (ST)-(S-T). So, it will be 779. So, the answer will be 959 indeed. Thanks for making it clear!

Mani Jha - 6 years, 1 month ago

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@Mani Jha But Since T2S2T^2-S^2 is a negative number, dividing by 1000 will give remainder -959. Isn't it?

Mani Jha - 6 years, 1 month ago

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@Mani Jha Wait. the answer will be 41 indeed. We don't want negative remainders. Suppose we want to find the remainder when -1959 is divided by 1000. While doing the normal division we do, we multiply 1000 by -2. What remains is 1959+2000=41-1959+2000=41. It can also be thought in this way: 1959=412000-1959=41-2000

1959(mod1000)=(41)(mod1000)(2000)(mod1000)-1959(mod 1000)=(41)(mod 1000)-(2000)(mod 1000) =41=41 So, answer must be 41 for the case at hand too

Mani Jha - 6 years, 1 month ago

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@Mani Jha Good, I can see that you are beginning to unwind several notions that are involved in this question.

At the end of the day you should be able to explain why one answer is correct, and the reason why you thought the wrong answer was correct.

Calvin Lin Staff - 6 years, 1 month ago

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But S2S^2 is divided evenly by 10001000 (you can stretch that to 10,00010,000, since 100100 divides SS, but that is unneccesary), so we really only need the remainder of T2T^2, which is easily checked to be 4141. Right?

Arthur Mårtensson - 6 years, 1 month ago

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I am not asking how to solve this problem; you can view the solutions to see how others approached it.

I'm asking for explanations of why the stated solution was wrong, and what misconception / assumption was made. How can we avoid such mistakes in future?

Calvin Lin Staff - 6 years, 1 month ago

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The idea behind keeping the last 3 digits is to find the remainder after dividing by 1000, i.e. mod 1000. However the last 3 digits are not (mod 1000) for negative numbers.

(T+S) is positive, so its (mod 1000) is the same as its last 3 digits, 421.

(T-S) is negative, so its (mod 1000) is not the same as its last 3 digits but rather 1000 - 779 = 221.

Continuing with the method we get:

421×221=93041421\times 221=93041 with last 3 digits 41.

Louie Tan Yi Jie - 6 years, 1 month ago

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you did it correctly then why is it coming different

SHASHANK GOEL - 6 years, 1 month ago

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