How do I solve $3(2^{x+2}-2^x) = 4a_1a_2a_3$

Question

I encountered this problem but I'm not sure how to solve it since it has 4 unknowns. $$3(2^{x+2}-2^x) = 4a_1a_2a_3$$

What is known is that $x\in\mathbb{Z}$ and $a_1, a_2$ and $a_3$ are digits in a 4-digit-number. I'm not even sure if a solution exist. If I divide by three I get:

$$(2^{x+2}-2^x) = \frac{1}{3}4a_1a_2a_3$$ So that's not working...

@EnthusiasticStudent I should find both $a$ and $x$. $a$ are unknown digits of the number on the right side. — Frank Vel, Jan 30 '15 at 13:17
Note that $2^{x+2}-2^x$ can be written as $2^x(2^2-1)$, which is $3\cdot2^x$. — MPW, Jan 30 '15 at 13:17
@MPW Then I get $x = \dfrac{\log_2(4a_1a_2a_3)}{2\log_2(3)}$ — Frank Vel, Jan 30 '15 at 13:22

score 2 · Answer 1 · answered Jan 30 '15 at 13:19

2

$$3(2^{x+2}-2^x) = 4a_1a_2a_3$$ $$3 \cdot 2^x \cdot (4-1)=4a_1a_2a_3$$ $$9 \cdot 2^x=4a_1a_2a_3$$

With a little of try and error you would see that $9\cdot 2^9=9\cdot512=4608$. Therefore:

$$x=9$$ $$a_1=6$$ $$a_2=0$$ $$a_3=8$$

answered Jan 30 '15 at 13:19

rlartiga

4,205
1
14
24

Is there no way without trying and erroring? – Frank Vel Jan 30 '15 at 13:20

score 2 · Accepted Answer · answered Jan 30 '15 at 13:35

I assume you want integer solutions.

You have that $3\cdot3\cdot2^x$ must be an integer between $4000$ and $4999$. So: $$4000\leq9\cdot2^x\leq 4999$$ $$\log_2\tfrac{4000}{9}\leq x\leq \log_2\tfrac{4999}{9}$$ which gives approximately $$8.795\leq x\leq 9.117$$ The only integer in this range is $$\boxed{x=9}$$ so that $$9\cdot2^x=9\cdot2^9=4608$$ and you have $$\boxed{a_1=6, a_2=0, a_3=8}$$

How do I solve $3(2^{x+2}-2^x) = 4a_1a_2a_3$

2 Answers2