AAAAAAAAAAAAAAAA Man, shame on my gray-haired ... I was just afraid that April Fool's jokes would creep into this thread in the form of unsolved problems and the like. Sorry
There is such a resource http://www.matri-x.ru/ There are also guys on their own wave and a lot of problems, though nobody has solved them yet, but they can twist your mind.
2 mod 2=0 12 mod 2^2=0 Take the number 12 (X) n=2;
X mod 2^n=0 X=K*2^n; add to it the most significant digit 1 or 2 ((1 or 2)*10^n+X) / 2^(n+1) ((1 or 2)*2^n*5^n + K*2^n) / 2^(n+1) ((1 or 2)*5^n + K)/2 if K is even, we take 2, if it is odd, we take 1 and get an integer
((1 or 2)*10^n+X) mod 2^(n+1) =0
Adding to the number which is divisible by 2^n either 1 or 2 in the capital, we get the number divisible by 2^(n+1). ......
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Man, shame on my gray-haired ...
I was just afraid that April Fool's jokes would creep into this thread in the form of unsolved problems and the like.
Sorry
A simple maths problem:
-
Give an example of a 3-digit natural number whose product of all 3 digits is 12 times their sum.
Next:
A simple maths problem:
-
Give an example of a 3-digit natural number whose product of all 3 digits is 12 times their sum.
Assuming that the digits are equal, then 66612 mod 2^2=0
Take the number 12 (X) n=2;
X mod 2^n=0 X=K*2^n;
add to it the most significant digit 1 or 2
((1 or 2)*10^n+X) / 2^(n+1)
((1 or 2)*2^n*5^n + K*2^n) / 2^(n+1)
((1 or 2)*5^n + K)/2
if K is even, we take 2, if it is odd, we take 1 and get an integer
((1 or 2)*10^n+X) mod 2^(n+1) =0
Adding to the number which is divisible by 2^n either 1 or 2 in the capital, we get the number
divisible by 2^(n+1).
......
Assuming the numbers are equal, then 666
Yes, that's exactly the number I was referring to.
Richie, will you solve it yourself?
So I came up with it myself :)))