How to find the multiplication of $pq \times abc$ such that the result is producing the same digits from the original problem?

Question

For example:

$$65 \times 281= 18265$$

$$65 \times 983= 63895$$

$$72 \times 936= 67392$$

$$87 \times 435= 37845$$

In general: the original figures reappear in the results of each of these multiplications.

How can we find another multiplication like these?

Is there any easy way to find the multiplication like these or Is there any theorem that state this unique multiplication?

Thanks

Answer 1

Here is the answer that I got by brute force using Haskell. You are welcome to try and make some sense of it. Note that the code below only counts cases where the product has no leading zeroes, i.e. is a number from $[10000, 99999]$. Also, it requires that each digit occurs the same number of times on the left and on the right of the "$=$" sign.

Prelude Data.List Control.Monad> let ans = [(x, y) | x <- [10..99], y <- [100..999], sort (show x ++ show y) == sort (show (x*y))]
Prelude Data.List Control.Monad> forM_ ans ( putStrLn . show )
(14,926)
(15,705)
(15,930)
(21,501)
(21,600)
(21,870)
(21,906)
(21,915)
(24,651)
(26,401)
(27,810)
(30,501)
(30,510)
(32,926)
(35,401)
(35,410)
(35,725)
(36,936)
(41,323)
(41,350)
(42,678)
(47,371)
(47,542)
(50,251)
(51,201)
(51,246)
(51,300)
(53,635)
(54,846)
(57,834)
(59,845)
(60,201)
(60,210)
(63,585)
(63,855)
(65,281)
(65,641)
(65,704)
(65,875)
(65,983)
(68,926)
(72,936)
(75,231)
(75,906)
(78,624)
(80,473)
(80,860)
(81,225)
(81,270)
(84,141)
(84,546)
(86,251)
(86,800)
(87,210)
(87,435)
(89,482)
(90,351)
(93,150)