Question 156377
Let's try to find the number of combinations that 3 and 7 can make up for a two digit number.


Let's try a smaller example. So let's use the numbers 3 and 7 to make a two digit number, three digit number, and a four digit number to see what is going on



So let's list ALL of the possible two digit numbers:


33
37
73
77


So for a two digit number, there are 4 possible combinations



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So let's list ALL of the possible three digit numbers:


333
337
373
377
733
737
773
777


So for a three digit number, there are 8 different combinations.

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So let's list ALL of the possible four digit numbers:


3333
3337
3373
3377
3733
3737
3773
3777
7333
7337
7373
7377
7733
7737
7773
7777


So for a four digit number, there are 16 different combinations.



Notice how the number of combinations keeps doubling (since 2 doubles to 4, 4 doubles to 8, 8 doubles to 16, etc...). 


So if we write out a 5 digit number only using 3 and 7, then we will list out 32 different combinations (which is double of 16). And finally for a six digit number, there will be 64 different combinations (which is double of 32). 



Note: the quicker way (ideally once you understand what's going on) is to use this formula {{{y=2^x}}} where {{{x}}} is the number of digits and {{{y}}} is the number of unique number combinations.


So in this case, {{{x=6}}} and {{{y=2^6=64}}}. So once again, we get 64 different combinations.



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Answer:


So there are 64 unique 6 digit numbers that can be generated from just the two values 3 and 7.