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A client buys 72 turkeys. On the receipt, 2 digits are missing, the first and the last. So, the total price is $\${-}67.9{-}$. Find a way to make the price of each turkey round at the second digit without trial-and-error method.

So, the answer is obviously $\$5.11$/turkey, so $\$367.92$ total. I found that with with 2-3 trial-and-error.

I bypassed his problem by using python and made a script that tries every combination and assures it is divisible by 0.01.

Please help me find a mathematical way of solving this problem.

Xander Henderson
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PyThagoras
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  • $!\begin{align} &\bmod 8!:\ 0 \equiv a679b \equiv790+b \equiv 6+b\Rightarrow b\equiv 2\Rightarrow \color{#c00}{b = 2},,\ \text{by },\ 0\le b\le 9\ &\bmod 9!:\ 0 \equiv a679b \equiv a!+!6!+!7!+!9!+!b\equiv a!+!\color{#c00}b!+4 \equiv a!+!6\Rightarrow a\equiv 3\Rightarrow a=3\end{align}\ \ $ – Bill Dubuque Dec 10 '20 at 18:26
  • We used casting out nines in the prior line, and in the first line we used that $,\color{#c00}{8\mid 1000},,$ by $,2\mid 10\Rightarrow 2^3\mid 10^3,,$ so $\bmod \color{#c00}8!: a6(\color{#c00}{1000})+79b\equiv 79b\ \ $ – Bill Dubuque Dec 10 '20 at 18:32

2 Answers2

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You need the divisibility rules for $8,9$ because a number is divisible by $72$ only if it is divisible by $8$ and $9$ as well.

According to that the rule, the sum of the digits must be divisible by $9$.

$$x+6+7+y = 22+x+y$$

In order to be divisible by 9, that sum must either be 27 or 36. The blanks must sum to $5$ or $14$.

It must also be divisible by $8$ so the last digit must be $2$. That means the first digit has to be $3$, because it would be impossible to make a sum of $14$.

Hence, the answer is x=3 and y=2

PyThagoras
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Ross Millikan
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Without loss of generality, we can multiply the prices by 100, to work with integers and avoid the commas. Let $n$ be the number of turkeys bought and $p$ be the price of each of them. Then $$72p=10000a+679*10+b$$ where $a,b$ are between $0$ and $9$. The trick is to notice that $72=9*8$. We can now use two divisibility tricks. If a number is divisible by $8$, then the last 3 digits must be divisible by it (this follows by the fact that $1000$ is divisible by $8$). Hence $79*10+b$ must be divisible by $8$. The only value of $b$ which satisfies this condition is $2$. The second trick concerns divisibility by $9$: if a number is divisible by $9$, then the sum of the digits is divisible by $9$. Hence we must have $$a+6+7+9+2=24+a$$ to be divisible by 9. The only value of $a$ which satisfies this condition is $3$. To find the price, we calculate: $$p=36792/72=511$$ Dividing back by 100, we get $p=5.11$ as you said.

A-B-izi
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