What are the possible solutions to 33⋅3333⋅33, if 11⋅11=411⋅11=4 and 22⋅22=1622⋅22=16?

Question

Elsie Hood

Mathematics

16 May, 03:25

What are the possible solutions to 33⋅3333⋅33, if 11⋅11=411⋅11=4 and 22⋅22=1622⋅22=16?

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Pete · Answer 1

Since 11*11 = 4, therefore in this case 11 = 2. Since 22*22 = 16, therefore in this case 22 = 4. So we can see that the common ratio is 2. Based from this, we can say that 33 = 6.

So that:

33*33 = 36