Calculator Mistake?

Number Theory Level 3

I bought four items from my local convenience store. The bill came to $7.11. I thought that amount must have been wrong, as I had seen the cashier multiplying the four prices together instead of adding them. (The cash register was broken, so he had to use a calculator instead.)

When I questioned the cashier about it, he agreed that he had multiplied the prices together, but said that in this case it didn't matter since the sum of the four items was also $7.11.

What was the price of the most expensive item (in dollars)?

Note: Each item costs a whole number of cents. For example, $2.11 is 211 cents, which is a whole number.

$3.16 $3.40 $3.26 $3.25

3 solutions

Sudarshan K.B
Mar 21, 2014

of everything i case of an MCQ the number which is a factor of 711 is the answer . So the answer is 3.16

Shourya Pandey
Mar 17, 2014

Converting everything to cents, we have $a+b+c+d=711$ and $abcd=711000000$ . Since $a, b, c, d$ are integers, they must be factors of $711000000=2^6*5^6*3^2*79$ . Of the given options, $325= 5^2*13 , 340=2^2*5*17 and 326= 2*7*23$ are not factors of $711000000$ , but $316=2^2*79$ is. So for an MCQ test , this method easily shows that $\$3.16$ is the answer.

Abhishek Sinha
Mar 17, 2014

Let the prices be $a \geq b \geq c \geq d$ , all of which are non-negative. We are given ,

$a+b+c+d= 7.11$ and $abcd= 7.11$ .

Applying AM-GM inequality on the non-negative variables $b,c,d$ , we have $\frac{b+c+d}{3} \geq (bcd)^{1/3}$

Substituting for $b,c,d$ , we have

$\frac{7.11-a}{3} \geq (\frac{7.11}{a})^{1/3}$

From which it can be easily verified that $0.74416\leq a \leq 3.19212$ are the non-negative solutions for $a$ .

How does that tell you what $a$ is?

I'm assuming that each of the prices are an integer number of cents. With that, I guessed that $a = 3.16, b = 1.5, c = 1.25, d = 1.2$ , but do not know if/why this is the unique solution.

Calvin Lin Staff - 7 years, 2 months ago

If you just look at the possible options for this multiple choice question, $a=3.16$ is the only choice. To investigate uniqueness, we have to look at the integral solutions of the following simultaneous diophantine equation $a+b+c+d=711, abcd=711*10^6= 2^6*3^2*5^6*79$ . Seems to be a tedious job !

Abhishek Sinha - 7 years, 2 months ago

I was afraid it was a tedious job + not necessarily unique solution. Let's see what others think.

Calvin Lin Staff - 7 years, 2 months ago

@Calvin Lin – I don't understand what do you mean by uniqueness , as the problem is to find the maximum value of $a$ over all (possibly multiple) such valid (meaning, integral in cents) solutions. Since you have showed $316$ cent is already a valid solution, you just need to check that $317,318,319$ are not factors of $711*10^6$ (whose factorization I have written above). This definitely shows that maximum valid a is indeed $3.16$ .

Abhishek Sinha - 7 years, 2 months ago

@Abhishek Sinha – I interpreted the question as saying that there is a set of $a, b, c, d$ which satisfies $a+b+c+d = 711, abcd = 711*10^6$ , and we are asked for $\max(a, b, c, d)$ .

In this sense, I am asking if the quadruple that I listed is the unique answer (without writing a computer program).

Calvin Lin Staff - 7 years, 2 months ago

Or, keeping in view the solution that you gave, you can just check that 317,318 or 319 are not a factor of $711*10^6$ .

Abhishek Sinha - 7 years, 2 months ago

Calculator Mistake?

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