https://nrich.maths.org/1130

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https://nrich.maths.org/1130

Quick question. My 8 year old daughter just got this homework. I would appreciate it, if an adult can solve it and tell me how long it took. I don't want your answer just how long.

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>>7817607

I'm a maths major, so don't take this as how long it should take your daughter or you, obviously. It took me about 5 minutes laying in bed, doing it in my had. I checked afterward to make sure my answers were correct as well. With a code, I could do this in a minute flat. It's definitely easier to do as a brute force problem, because any other way requires a level a critical thinking that is outside of most people's comfort zones.

It's a good problem in my opinion, and it encourages a very nice level of critical thinking, but it's definitely not for an 8 year old to do alone. This is something to would expect to see early on in a high school/undergrad level maths logic class as a solo homework question.

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>>7817607

7 minutes. this is a nice problem.

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Thanks guys.

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>>7817607

12 minutes, thinking about it while doing something else and without writing anything down (which wouldn't actually have helped much).

It's really not that hard once you figure out some obvious boundaries. Hope your wife's daughter got it right.

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>>7817607

Took me 3 minutes. Would be hard for an 8 year old and they would be bruteforcing it if they could understand it.

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just below 10 mins

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2 minutes, the way it's written out for you makes this really easy. Congrats on sending your daughter to a school where she isn't treated like a fucking retard though.

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>>7817695

>Would be hard for an 8 year old and they would be bruteforcing it if they could understand it.

This. It was easy, but I had to use an equation for it.

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Like 2 minutes, tbqhfam

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>>7817744

You only found one though. I thought the exercise was finding them all.

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>>7817607

Its pretty easy if youre into university level maths. Probably 10 minutes.

Its just unfair to make an 8 year old do that. Im pretty sure the average adult couldnt solve this.

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>>7817758

Oh whoops, I misread it. Time to commit sudoku.

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>>7817777

One part is two find a solution. It's the plus part to figure out how many you can make.

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>>7817805

Could anyone say by the next 14 hours or however long this thread remains open how many possibilities there are?

It took me 11 minutes. I did not use an equation and was slightly distracted, but I am afraid I could not have done much better than this.

Did not use equations, just looked at how I can make sure that some parts fit so the task gets easy... and even though I found out how to do it after ~200 seconds it took me this long.

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>Your challenge is to find four different digits that give four two-digit numbers which add to a total of 100.

>How many ways can you find of doing it?

Wait, were people just looking for four solutions or for the total number of possible solutions? I worked out the total number possible, it didn't take me too long (somewhere between 10 and 15 minutes) but definitely longer than 2 minutes.

>>7817744

I did this but took it further. Though later on when I was almost finished I realized an easier way to trivialize it through intuition.

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>>7817829

I am the one you responded to, and I found more than that.

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>>7817834

Another anon here, used linear algebra and also concluded that there are only 4 solutions

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>>7817607

First year comp sci took me sub five minutes. Also I didn't write a fucking python script like an autist would before any of you fucks accuse me of. I'm just giving OP context of my abilities.

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>>7817842

You are not even close anon, and I'm not trying to be mean.

The answer is in the double digits.

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>>7817849

but that's wrong and it's easy to show

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>>7817853

Go ahead anon. Impress us with your abilities.

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>>7817849

Are you sure you read the question correctly?

All numbers must be distinct and between 1 and 9.

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>>7817805

>two find

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>>7817834

I'd like to see your answers then.

Let me explain my reasoning:

20a+11b+11c+2d = 100

"the ones place numbers" = b+c+2d = 10,20,30

"the tens place numbers" = 2a+b+c = (100-the ones place numbers)/10 = 9,8,7

2d-2a = 1,12,23

d = 0.5+a, 6+a, 11.5+a

Only d = 6+a results in an integer and there are only 3 possibilities

a = 1, d = 7

a = 2, d = 8

a = 3, d = 9

b+c = 20-2d = 6,4,2

Since the numbers must be different (forgetting about this is probably where you got the extra answers) the only possibilities are

a = 1, b = 4, c = 2, d = 7

a = 2, b = 3, c = 1, d = 8

And b and c are interchangeable so this gives us 4 answers.

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>>7817863

Oh, I didn't realize they had to be distinct. That changes things. I'm pretty sure it's still trivially more than 4 but I do need to redo my computations. Unfortunately I'm headed out the door so I can't do it right now.

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>>7817865

Typo. English isn't my native tongue, I typed too quickly.

>to

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>>7817868

>misreads question

>"oh I didn't do the math but pretty sure you are still wrong"

What's the matter can't do it under 5 mins?

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>>7817744

>senpai

why havent you killed yourself

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>>7817885

the question says the digits have to be from 1-9. c can't be 0. Reading the problem correctly is the first part of solving it anon :)

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>>7817879

Just fuck off with your arrogance.

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>>7817892

>can't solve math problem for 8 year old girls

>calls people who can arrogant

By the way I wrote out the same equation as >>7817867 but then I just looked at the possibilities for b and c together (you quickly get just [1, 3], [2, 4] and maybe [1, 5] if you don't yet see it's not possible) and then brute force the rest.

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>>7817899

Oh, the cancer reached this thread too.

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>>7817829

>>7817835

Not the same guy, but I also got 4.

Let's call the numbers in each cell:

A B

C D

The final sum will be 20A + 11B + 11C + 2D.

There are no 0s, so obviously A can't be bigger than 4. Then it's just about trying A's possible values and seeing which products of 11 + an even number no bigger than 18 add up to the remainder of 100.

A = 4

20A = 80, there's no way to make something smaller than the remaining 20 with the other factors.

A = 3

20A = 60, you have 40 left. If B = 1 and C = 2, 11(B + C) = 33, you have 7 left for 100. But 2D can't make an odd number. And if you make B or C any bigger, the sum is already greater than 40, so this can't be it.

A = 2

20A = 40, 60 left. If B = 1 and C = 3, 11(B + C) = 44, 16 left, therefore D = 8. This counts as 2 solutions, because you can also make B = 3 and C = 1 with the same result. Making B and C any bigger becomes greater than 60.

A = 1. With 20A, you have 80 left. The biggest 2D can get is 18, therefore 11(B + C) must be bigger than 62. 66 is the only possible value for it, anything else is odd or too big. 1 is already in use, so B and C can only be 2 and 4, two ways to make it. D = 7.

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>>7817911

>1 is already in use, so it can't be used again

Irrelevant. In order to not repeat numbers, the only criteria is that b != c

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>>7817940

Huh? If a is 1 neither b nor c can be 1, even if b and c are not the same.

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Took no time at all

3 1

1 9

I just used the numbers they gave you

Realize that it needs to equal 100, they gave you 151

Changing the top left one by 1 will change the answer by twenty

Change the top right or bottom left one by 1 and it will change by 11

Change bottom right one by 1 and it will change by 2

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>>7817963

The digits must be different, read the problem and try again.

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>>7817886

I like seeing you mad, baka desu senpai, and no, I wasn't filtered this time :^)

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>>7817960

The way I was reading the problem, the four terms at the end have to sum to 100 and you can't have repeating terms. Nothing to say that your individual numbers can't repeat though.

Take, for example, the case of a=1, b=1, c=5, and d=7. Then your four terms would be 11, 57, 15, and 17, which adds up to 100. Did I read it wrong?

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>>7818323

Whoops. Going back and reading it again, it looks like I did read it wrong. My mistake.

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>>7818323

>Your challenge is to find *four different digits* that give four two-digit numbers which add to a total of 100.

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It took me like 5 minutes before I realized that I was adding the numbers incorrectly, then it took me like 30 seconds.

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5 seconds

t. a 145 iq math monk

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12

47

3mins with a calculator

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>>7817744

>How many ways can you find of doing it?

You need to find out how many solutions exist, you dolt

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>>7819954

I don't think thats the case. It is probably just a comment to get the smart kids thinking. There is no way that the question is actually to find how many possible combinations are there.

Do you remember how much math you were taught when youre 8? You wouldn't know where to start. How would you prove that your answer is correct?

The best thing an 8 year old can do is bruteforce it and no 8 year old has that amount of concentration.

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There are 4. Didn't take me long. Basically only used addition and logic....excuse the handwriting.

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8 y/o math at its finest.

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>>7817607

I'm pretty tired after a long day of work, but it took me around 2 minutes. With pencil and paper I'd do it in 30 secs tho. It's a great problem for an 8 year old, but I would have had a very hard time figuring it out when I was 8 I think. I dont think I had ever seen symbolic math by that time, so unless I had a really big incentive I can't imagine solving it at all, and even if it did it would have taken me a long while. Pretty brutal for a homework assignment unless your daughter is a genius.

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