# Project Euler #32: Pandigital products

# Project Euler #32: Pandigital products

cnhnyu + 1 comment For N = 9, I got the following result:

4*1738=6952

4*1963=7852

12*483=5796

18*297=5346

27*198=5346

28*157=4396

39*186=7254

42*138=5796

48*159=7632

sum = 56370

I passed all tests except the last one. Can someone please tell me what's going wrong here?

cnhnyu + 1 comment Got it. No duplicate product. Not no duplicate multiplicand/multiplier pair.

bhavikgevariya + 0 comments [deleted]

coder_batman + 0 comments what is the upper limit of the numbers to be considered for multiplicand and multiplier ?

tushartyagi8750 + 1 comment i am getting error for numbers greater than 7 i am using the permutation function.any help will be appericiated

prunns + 0 comments I used the same thing. At the beginning I was also getting wrong answers for TC 1,3 and 5. What kind of error are u getting?

I could pass all the cases with a little modification in pypy3 with the last case taking 1.21s

deekshit_842 + 0 comments ive used brute force approach to solve this and got timeout in testcase 5 can anyone explain it

debagnikroy + 1 comment i am getting sum as 0 for n=7. is it correct? my test case 4 is only failing

tushartyagi8750 + 0 comments no it's not 0 for 7

allam_hemanthku1 + 0 comments what we have to add to get the sum becase i am simply breaking the loop .

Haseena_CSE + 1 comment Could you please tell me what is the logic here?

SOKS33 + 0 comments That's mine :

Generate all pandigital numbers of provided size (I use string permutation to compute them quite fast)

Then split into a/b/c of various sizes and try to find a match in a * b = c

Save c only once in your sum (ex : 4312 and 3412 give 4*3 = 12 and 3*4 = 12. Save 12 only once !)

jenilvagadiya19 + 0 comments nicely done Let's say i*j=k is pandigital then for N=5 i=1 j=2 k=2 (number of digits) N=6 i=1 j=2 k=3 and so on ... upto N=9

god_blessme + 1 comment Is there any way other than brute force??

SOKS33 + 0 comments In pandigital generation yes there is better. String permutation ("1234" gives "1234" & "1243" etc.)

abdghani + 1 comment please see my test case 5 is failing..

cantonios + 0 comments It's because they say "Some products can be obtained in more than one way so be sure to only include it once in your sum."

they mean more than "with the same multiplicand/multiplier"... if you get the same product with different multipliers, that should be excluded as well.

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