#Math:
35! = 10333147966386144929*66651337523200000000
ga teng. Qo'yida keltirilgan raqamlardan qaysi biri * o'rniga kelishi mumkin

PS:Computerdan foydalanmaysilar degan umiddaman

#Math_Force
Game va Animation productive dasturlashdagi matematikaning o'rni albatta oxirigacha ko'rishni tavsiya etaman.

Bir ajoib va menimcha hammaga tezda tushunarli buladigan masalani sizlar bilan bo'lishmoqchiman.

Menimcha hamma faktorial haqida xabari bor.
n! (yoki n faktorial) bu 1 dan n gacha bo'lgan barcha sonlar ko'paytmasi.
Masalan:
4!=1*2*3*4
5!=1*2*3*4*5
........................
Demak n! = 1*2*3*4*....*n (0!=1 deb olishadi odadta)
va albatta hamma raqamdan sonni farqiga boradi degan umiddaman.
___________________________________________________________________________
Agar doskada d raqamini n! marotaba yozsak u qaysi toq raqamlarga bo'linadi?(Bu codeforces sodda masalalaridan biri)
n = 1...10^9 gacha
Masalan d=1 n=3
3!=1*2*3=6
111111-shu son qaysi toq raqamlarga bo'linadi?
toq raqamlar 1 3 5 7 9
shu lardan 1 3 7 ga bulinadi.
Algoritmingizni izohlarda qoldiring.

#Logic
#Math

en:
I want to share with you an interesting problem that I think will be easy for everyone to understand.

I believe everyone is familiar with the concept of a factorial.
n! (or n factorial) is the product of all numbers from 1 to n.
For example:
4! = 1 × 2 × 3 × 4
5! = 1 × 2 × 3 × 4 × 5
...
So, n! = 1 × 2 × 3 × 4 × ... × n (and by convention, 0! = 1).
Hopefully, everyone understands this so far.

The problem:
If the digit d is written n! times on the board, which odd numbers will it be divisible by?
(This is one of the simple problems from Codeforces.)

Given:
n = 1...10^9
Example:
d = 1, n = 3
3! = 1 × 2 × 3 = 6
Writing 1 6 times gives us 111111. Which odd numbers is this divisible by?

Odd numbers: 1, 3, 5, 7, 9
From these, 111111 is divisible by 1, 3, and 7.

Leave your algorithm in the comments!
#math
#logick