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A student made an error while multiplying two three-digit numbers. Instead of noticing the multiplication sign between them, the student wrote the numbers together, forming a six-digit number.
Surprisingly, this mistake resulted in a product that was three times greater than the actual product of the two numbers.
Can you find these two original three-digit numbers?
A = 167, B = 334.
We start by assigning variables to the two three-digit numbers: let A represent the first three-digit number and B represent the second. According to the given condition, if we mistakenly combine these numbers, forming a six-digit number, we get 1000A + B. Surprisingly, this erroneous expression equals three times the actual product of the two numbers, which gives us the equation 1000A + B = 3AB.
Rearranging this equation, we get (3A - 1)B = 1000A. Since A and 3A - 1 are coprime, we deduce that 1000 must be divisible by 3A - 1.
Considering the possible values for 3A - 1, we recognize that it must be a three-digit or four-digit number since A is a three-digit number. Therefore, 1000 has two potential divisors fitting this criterion: 500 and 1000.
However, 3A - 1 cannot equal 1000 because 1001 (which is 3A) is not divisible by 3. Thus, we deduce that 3A - 1 must equal 500. Solving this equation, we find that A equals 167. Substituting this value back into the original equation, we can solve for B, which equals 334.
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