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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.

Alauna Gentz
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  • Let the missing digits be $a,b,c$. So that 2._ ends with only one digit and not several, you need that $a\times b$ be a multiple of ten, as you correctly noted. That directly implies that one of $a,b$ must be a multiple of five and one of $a,b$ be a multiple of two. You can rule out $a$ being $0$ or $5$ since $0.25\cdot 7.9 < 2$ and you can rule out $b$ being zero since $0.28\cdot 7.0<2$ and the only other possibility here would have been $0.29\cdot 7.0=2.03$ which does not end in only one digit. It follows that $b=5$ and $a$ is even. $2.6\cdot 7.5<2$ leaving $a=8,b=5$ as only possibility – JMoravitz Apr 25 '22 at 14:50

2 Answers2

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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.

user376343
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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.

Gareth Ma
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