Place a nonzero digit in each space so that the equation is true.
0.2_ * 7._ = 2._
Here is the work I've done so far:
2/7=0.285714...
Then I did some guessing and checking and got 0.28 * 7.5 = 2.1. This works I just need to know if there is smarter way to do this other than guessing and checking.
I also know I need the hundredth's place * the tenths place to have a base of 10 so that the 2._ just has one number following.
On
We're given $0.2\_ \times 7.\_ = 2.\_$
We can remove the decimal places by multiplying $1000$ on each side. Then we get $2\_\times 7\_ = 2\_00$.
From here, notice that RHS is divisible by $100$, which means it's divisible by both $4$ and $25$. In particular, one term is divisible by $5$, which gives $4$ cases:
$2\_ = 20$ -> $20\times 7\_ = 2\_00$, clearly impossible by dividing RHS by $20$.
$2\_ = 25$ -> $25\times 7\_ = 2\_00$, which is still impossible, since $\frac{RHS}{25} = \frac{2\_00}{25}\geq \frac{2000}{25} = 80$.
$7\_ = 70$ -> $2\_00 = 2100$ or $2800$ are the only multiples of $70$ and $100$, but that means $2\_ = 30$ or $40$.
$7\_ = 75$ -> $2\_ \times 75 = 2\_00$. As you have found, $28\times 75 = 2100$ is the only solution.
Hope this helps.
Multiplying the equation $$0.2\_ \times 7.\_ = 2.\_$$ by $1000$ makes us to work with natural numbers. Denote $a,b,c$ the missing digits in this order. The task is to solve the diophantine equation $$(20+a)(70+b)=2000+100c$$ for $a,b,c\in\{1,2,\dots ,9\}.$
$RHS$ is divisible by $100,$ hence $a$ or $b$ equals $5.$ As $25$ (obtained for $a=5$) and $75$ (for $b=5$) are divisible by $25,$ the other missing digit is even, and the corresponding sum is divisible by $4.$
With $a=5$ is $LHS<2000<RHS,$ therefore $a\neq 5.$
We have necessarily $b=5.$ The only values of $a$ such that $(20+a)$ is divisible by $4,$ are $4$ and $8.$ As $$24\times 75=1800<2000<RHS,$$ we conclude $a\neq 4.$
The missing digits are $a=8,b=5$ and $c=1,$ obtained after uploading $a,b$ into the equation.