The discussion on this point is accepted with open arms in the community; the community answers the question with proof of the existence of certain other numbers on a variety of numerics.

a = 7, b = 5 , n = 5 175 = 1 x 7 x 5 x 5

a = 2, b = 8, n = 8 128 = 1 x 2 x 8 x 8

a = 3, b = 5, n = 9 135 = 1 x 3 x 5 x 9

a = 4, b = 4, n = 9 144 = 1 x 4 x 4 x 9

Starting with the acceptance that several digits can exist in the pattern discussed in the problem proves an abundant variable available that you can find. Though there was some difficulty in understanding the equation, the community did find the solution keeping the context that the number can be in any range and giving the result lies the numbers in the statement.]]>

3 x 6 times [letter x] = 36 => 18x = 36 => x = 36/18 = 2

Bob

]]>Welcome to the forum.

I think that there may be many such. This is how I'm going to start looking.

You seem to be happy to multiply by 9 or 14 etc so I'll assume all you want is [number] = [any multiple of] [product of digits]

So algebraically:

Find a b and n so that 10a + b = nab for whole numbers a, b and n. {this is the two digit version}

I'll see if I can find some more .

On my spreadsheet allowing a and n to be from {2,3,4,5,6,7,8,9} I only got one more: b = 4 and 24 = 2 x 4 x 3

Three digits may take longer.

Actually not much longer as it was quick to add 100 to my 10a + b.

I found four in the one hundreds:

a = 7, b = 5 , n = 5 175 = 1 x 7 x 5 x 5

a = 2, b = 8, n = 8 128 = 1 x 2 x 8 x 8

a = 3, b = 5, n = 9 135 = 1 x 3 x 5 x 9

a = 4, b = 4, n = 9 144 = 1 x 4 x 4 x 9

Bob

]]>So 36 = (3x6)x2

Is there any other number that can duplicate this? Namely ABCD--n digits = (AxBxCxDx--n digits) x2? If not why not? I found other numbers with different multipliers like:

144 = (1x4x4)x9 and 224 = (2x2x4)x14

Are these unique numbers?

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