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Forum ยป Misc IQ test (900+ IQ only)
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Thread: Misc IQ test (900+ IQ only)

post 1684391713 06-10-2023, 10:13 PM
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Misc IQ test (900+ IQ only)

Find the number of sets of 3 distinct positive integers whose product is 69350.

Gave a ****** 9th grader this kind of problem and he solved it in a minute. Should be no problem for you white super high IQ elite ubermenschen. Let's hear it.
post 1684391743 06-10-2023, 10:14 PM
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Bout 350.
post 1684391803 06-10-2023, 10:15 PM
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Originally Posted By ElonFan96⏩
Bout 350.
depending on what u mean by bout u are correct, but can u give the exact number bro

I wonder if ChatGPT can do it.
post 1684391853 06-10-2023, 10:16 PM
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1) 1, 2, 69347
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post 1684391913 06-10-2023, 10:18 PM
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Eleventeen.
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MAGA
post 1684391933 06-10-2023, 10:18 PM
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Originally Posted By BraddlesMcGee⏩
1) 1, 2, 69347
nope
Originally Posted By BigNik78⏩
Eleventeen.
nope

lul

ChatGPT also gets it wrong lol.
post 1684391963 06-10-2023, 10:19 PM
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Op-what percent of black Americans can do that in under a minute?
post 1684391993 06-10-2023, 10:20 PM
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Originally Posted By 129iq⏩
Op-what percent of black Americans can do that in under a minute?
irrelevant

question is can U solve it brah
post 1684392033 06-10-2023, 10:22 PM
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If n = 10

1, 2, 7
1, 3, 6
1, 4, 5
2, 3, 5

We need a formula f(n). It's not so hard.
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post 1684392063 06-10-2023, 10:23 PM
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wincel are u ****** or jew
post 1684392103 06-10-2023, 10:24 PM
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Originally Posted By BraddlesMcGee⏩
If n = 10

1, 2, 7
1, 3, 6
1, 4, 5
2, 3, 5
2, 4, 4


We need a formula f(n). It's not so hard.
Closed form for arbitrary n is quite hard actually...lmao but not bad with some number theory

I wouldn't bother with a closed form one. Try a dif approach using counting.
post 1684392123 06-10-2023, 10:24 PM
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just explain the answer for the other miscers. this was too ez for me
post 1684392133 06-10-2023, 10:25 PM
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Originally Posted By unusefulidiot⏩
Closed form for arbitrary n is quite hard actually...lmao but not bad with some number theory
I updated it because I repeated 4 in the last one.
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post 1684392173 06-10-2023, 10:26 PM
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My iq is 90. Just saying
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post 1684392203 06-10-2023, 10:27 PM
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25, ,38, 73
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post 1684392223 06-10-2023, 10:27 PM
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13870 x 5 x 1
post 1684392233 06-10-2023, 10:28 PM
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Originally Posted By DYELstatus⏩
just explain the answer for the other miscers. this was too ez for me
I'll explain it later. Want to see if some high iQ miscer gets it. Chryssipus, Numbersguy12, and Miscmathematician are not allowed.
post 1684392293 06-10-2023, 10:29 PM
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Originally Posted By paulinkansas⏩
13870 x 5 x 1
That's 1 such set. I want the number of all such sets. (And remember, order doesn't matter in a set so {1,2,3} is the same as {3,2,1} as a set.
post 1684392323 06-10-2023, 10:29 PM
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funnily enough this kind of problem is pretty common if you're in software engineering and do interview problems

it's essentially a 2 pointer type problem. just you have a 3rd one which is a fixed pointer

solution 1: 1, 2, 69347. then converge by incrementing the second number and decrementing the third number

solution 2: 2, 3, 69345. then converge

i could write a program that solves this in a minute.

not that much harder mathematically.
post 1684392383 06-10-2023, 10:32 PM
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Originally Posted By unusefulidiot⏩
That's 1 such set. I want the number of all such sets. (And remember, order doesn't matter in a set so {1,2,3} is the same as {3,2,1} as a set.
post 1684392393 06-10-2023, 10:32 PM
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Originally Posted By outfoxxed⏩
funnily enough this kind of problem is pretty common if you're in software engineering and do interview problems

it's essentially a 2 pointer type problem. just you have a 3rd one which is a fixed pointer

solution 1: 1, 2, 69347. then converge by incrementing the second number and decrementing the third number

solution 2: 2, 3, 69345. then converge

i could write a program that solves this in a minute.

not that much harder mathematically.
You could write a program to do it, but that's the stupid way. There's a fast way. Can you figure it out?
post 1684392413 06-10-2023, 10:32 PM
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OK product not sum.
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post 1684392423 06-10-2023, 10:33 PM
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Originally Posted By paulinkansas⏩
lmao wut
post 1684392433 06-10-2023, 10:33 PM
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Originally Posted By outfoxxed⏩
funnily enough this kind of problem is pretty common if you're in software engineering and do interview problems

it's essentially a 2 pointer type problem. just you have a 3rd one which is a fixed pointer

solution 1: 1, 2, 69347. then converge by incrementing the second number and decrementing the third number

solution 2: 2, 3, 69345. then converge

i could write a program that solves this in a minute.

not that much harder mathematically.
Doesn't product equal multiplication?
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post 1684392473 06-10-2023, 10:34 PM
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Originally Posted By Godfrd824⏩
Doesn't product equal multiplication?
yea he is doing sum, which is wrong, but you can write a program to do this problem easily

but I want to see if anyone can figure it out...it requires basic combinations and counting...

https://discrete.openmathbooks.org/d...-and-bars.html
post 1684392543 06-10-2023, 10:36 PM
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The prime integers are 2, 3 and 5. You just have to use the right combination of them to get to 69350.
post 1684392613 06-10-2023, 10:39 PM
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Originally Posted By paulinkansas⏩
The prime integers are 2, 3 and 5. You just have to use the right combination of them to get to 69350.
3 is not a factor. (Sum the digits.)

But you are on the right track looking at the prime factorization.

I'll go ahead and tell you the prime factorization is 2*5^2*19*73.
post 1684392693 06-10-2023, 10:41 PM
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Originally Posted By unusefulidiot⏩
3 is not a factor. (Sum the digits.)

But you are on the right track looking at the prime factorization.
Then it is 2 and 5. I haven't done this sort of math since 1987.
post 1684392843 06-10-2023, 10:46 PM
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Originally Posted By unusefulidiot⏩
3 is not a factor. (Sum the digits.).
Most people don't know if you sum the digits you can find if it is divisible by 3.
post 1684393023 06-10-2023, 10:52 PM
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Originally Posted By paulinkansas⏩
Most people don't know if you sum the digits you can find if it is divisible by 3.
That follows because 10 is 1 mod 3, and so are all powers of 10.
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