How many 10-digit numbers can be formed with product of digits equal to 2^{27}?

[Marcelo Arevalo]

How many 10-digit numbers are there such that the product of its digits is
equal to 2^{27}?

 

[MarkFL]

What must be true of all the digits?

 

[Marcelo Arevalo]

My son's solution:
please comment, thank you.

We can only use 4 digits : 1,2,4,8 (2^0 , 2^1, 2^2, 2^3)
if we use 9 8's
8-8-8-8-8-8-8-8 and 2^n
2^27 . 2^n = 2^27
2^n = 1

8,8,8,8,8,8,8,8,8,1 only possible digits if true are 9 8's.
_8_8_8_8_8_8_8_8_8 = 10 spaces to place the 1

10C1 = 10 ways to arrange the "1" . Thus, there are 10 of there numbers.

if we use 8 8's
8-8-8-8-8-8-8-8-x-y = 2^27
2^24 . xy = 2^27
xy = 2^3
since we only use 8 8's, we can't use 8.
(x,y) = (2,4) or (4,2)

_8_8_8_8_8_8_8_8_ _ spaces for numbers can be at the end.

since order matters : 10P2 . 10!/8! = 90 Ways if we use 7 8's
8-8-8-8-8-8-8 . abc = 2^27
2^21 . abc = 2^27
abc = 2^6
(a, b , c ) = (2^2, 2^2, 2^2)
= (4, 4, 4)

_8_8_8_8_8_8_8_ _ _ = 10 spaces
place all the three nos. can be at the end

10C3 = 120

we can no longer use 6 8's
since 8-8-8-8-8-8-4-4-4-4 \ne 2^27
2^36 \ne 2^27

Therefore : 10+90+120 = 220

 

[MarkFL]

Yes, I agree with your result. All digits must be a one-digit power of 2 (1,2,4,8).

If we use (8,8,8,8,8,8,8,8,8,1) we have:

$$N_1=\frac{10!}{9!}=10$$ ways to arrange.

If we use (8,8,8,8,8,8,8,8,4,2) we have:

$$N_2=\frac{10!}{8!}=90$$ ways to arrange.

If we use (8,8,8,8,8,8,8,4,4,4) we have:

$$N_3=\frac{10!}{7!\cdot3!}=120$$ ways to arrange.

So, the total number $N$ of such numbers is:

$$N=N_1+N_2+N_3=220$$