Problem 266

[MrDrake]

This is annoying... I've got 232...442, but there must be one larger!

EDIT: Never mind... there must be something wrong with my code somewhere, but changing a variable gave me the right answer!

[sivakd]

Not sure if asking this question would give away too much info, can this be solved with just 64 bit integers or like BigInteger type data types are needed?

[jaap]

That's a false dichotomy.

[sivakd]

Thanks jaap. I managed to solve this problem without requiring big ints.

[JMW1994]

What does this exactly mean?

Let p be the product of the primes below 190.

[thundre]

What does this exactly mean?

Let p be the product of the primes below 190.

p = 2 * 3 * 5 * 7 * 11 * ... * 181

[JMW1994]

Thanks.

[MYNick]

https://projecteuler.net/problem=266

Hey guys, im trying to solve problem 266 for a couple of days now. Im pretty sure my solution is correct but the result is big red x. So i think i might do something wrong and i didn't understand the question correctly.
The question is:
lets take num which equal to the product of the prime numbers until 190 (the product of 42 numbers).
Now we take the sum and make a square root out of it. Now i need to find a product of group of numbers from the primes list (the 42 primes). For example: the third, fifth and 10th numbers will be the product. And i need to find the one that is closest to the sqrt of the real number. BUT it must be the closest from below, i can not pass the sqrt.

In the end i mod this product by (10**16)
I tried it in serval ways and i have no idea why the result is wrong. Can some one confrim something for me?
For the primes number in 100 (25 primes) my result is: 9552415660183671 (after mod). can some one confirm please?
The problem with the examples in the question is that they all 1 prime product and not product of couple of primes. Would be helpfull if real example will be added.

[DJohn]

Your description sounds right to me (except in "Now we take the sum and make a square root out of it", it's the product, not the sum). N is the product of the primes below 190 (a very large number). You need to find the largest s such that s divides N and s <= sqrt(N).

It's easy to check small values by brute force. For primes below 20, I get N = 9699690 and s = 3094. For primes below 50 (the largest N that fits in 64 bit integers), s = 783152070. For the primes below 100, my result looks nothing like yours. Mine is 10xxxxxxxxxxxxxx65.

[MYNick]

Your description sounds right to me (except in "Now we take the sum and make a square root out of it", it's the product, not the sum). N is the product of the primes below 190 (a very large number). You need to find the largest s such that s divides N and s <= sqrt(N).

It's easy to check small values by brute force. For primes below 20, I get N = 9699690 and s = 3094. For primes below 50 (the largest N that fits in 64 bit integers), s = 783152070. For the primes below 100, my result looks nothing like yours. Mine is 10xxxxxxxxxxxxxx65.

for below 100 i received 15xxxxxxxxxxxxxx65 but i got the right reslt for 190 now. I didnt receive the same result like you for 50 and i found the problem with my code.
I used sort in the wrong way in python.
I now solved it, thank you very much

[v6ph1]

for below 100 i received 15xxxxxxxxxxxxxx65

As 2*3*...*97 = 2.3 * 10^36 and therefore the square is around 1.5(18..) * 10^18, DJohn must have a typing mistake.

[DJohn]

for below 100 i received 15xxxxxxxxxxxxxx65

DJohn must have a typing mistake.

Yes, I don't know how that happened. I get 15xxxxxxxxxxxxxx65 too.

I intended to give the value after the mod, but a) that doesn't start with 10 either, and b) I got the wrong number of digits. I'll just blame Friday afternoon.

[square1001]

I got Wrong Answer in my program.

Let f(n) = (the answer where p = (product of primes below n) ). (after taking mod)

f(30) = 79534
f(50) = 783152070
f(70) = 2803119896185
f(100) = 840xxxxxxxxxx365 (16 digits)
f(150) = 60xxxxxxxxxx310 (15 digits)
f(190) = 884xxxxxxxxxx635 (16 digits)

What value is wrong?
Please tell me.

[v6ph1]

The two last ones are definitely wrong - Did you take care about the range of 64Bit integers?

[square1001]

Thanks, v6ph1!
I realized that my code had been overflowed...
Finally I got AC