Posted - 05/17/2013 : 12:04:55 An oldie that you might have come across before.

A customer at a 7-11 store selected 4 items to buy, and was told that the cost was $7.11. He was curious that the cost was the same as the store name, so he inquired as to how the figure was derived. The clerk said that he had simply multiplied the prices for the 4 induvidual items.The customer protested that the 4 prices should have been ADDED, not MULTIPLIED. The clerk said that that was okay with him, but the result was still the same, exactly $7.11 What were the prices of the four items?

8 L A T E S T R E P L I E S (Newest First)

TchrWill

Posted - 05/24/2013 : 15:47:23

quote:Originally posted by the_hill1962

Yes, I like your explanation. The solution is, indeed $1.20, $1.25, $1.50 and $3.16 Now, how was this problem conceived? Can anyone create a problem such as this?

If you enjoy these puzzles, there are other numbers that work like this. $7.11 is not the only number which works. Here are the first 160 such numbers, preceded by a count of distinct solutions for that price. While $7.11 has a single, unique solution, some of the other numbers have several distinct answers. I found these many years ago but have lost the site address.

Posted - 05/23/2013 : 12:49:47 Yes, I like your explanation. The solution is, indeed $1.20, $1.25, $1.50 and $3.16 Now, how was this problem conceived? Can anyone create a problem such as this?

TchrWill

Posted - 05/22/2013 : 10:47:35 Here is my best shot at this.

<< A customer at a 7-11 store selected 4 items to buy, and was told that the cost was $7.11. He was curious that the cost was the same as the store name, so he inquired as to how the figure was derived. The clerk said that he had simply multiplied the prices for the 4 induvidual items.The customer protested that the 4 prices should have been ADDED, not MULTIPLIED. The clerk said that that was okay with him, but the result was still the same, exactly $7.11 What were the prices of the four items? >>

I have wrestled with other explanations of this oldie but, quite frankly, have found them difficult to understand. With the benefit of these explanations and some soul searching of my own, I believe I have derived a more easily understood solution.

The prime factorization of 711,000,000 is 2^6(5^6)3^2(79^1) which results in 294 factors/divisors making it impossible to take the trial and error approach in the hope of hitting on the desired 4 numbers. Therefore,

1--Converting the $7.11 to 711 cents, we are left with the problem of finding four factors of 711,000,000 that add up to 711. 2--In order to produce the six zeros in the final product, the result of any individual multiplication must end in zero, i.e., the six possible multiples of the four factors in pairs must all end in zero. 3-- All four factors cannot end in 5, as the sum of the unit digits of the four numbers must sum to 1, 11, 21, or 31. 4--To derive a 1 in the units place, two of the factors must end in 1-0, 2-9, 3-8, 4-7, or 5-6. 5--Only 1 and 0 or 5 and 6 will produce the zero last digit as required from step (2). 6--Therefore, the two possible sets of factors are of the form ...XX1..........XX5 ...XX0..........XX6 ...XX0..........XX0 ...XX0..........XX0 7--Some multiple of 79 must be one of the factors. 8--By taking the cube root of N = [711,000,000/79x], we obtain a reasonable estimate of the geometric mean of the other three factors and their sum with 79x.

9--Clearly, only 2, 3, or 4 times 79 appear to offer possibilities as the others produce sums greater than 711. 10--None results in a units digit of 1 meaning that our units digits must bx 0, 0, 5, and 6. 11--Only 4x79 = 316 ends in a 6 which we now know we need. 12--Therefore, we are reasonably confident that 316 is one of our four factors. 13--711,000,000/316 = 2,250,000 the cube root of which is 131+, roughly the geometric mean of our three remaining factors. 14--One of these three factors must end in a 5. 15--Lets try a range of candidates:

16--Clearly, only 125 results in an evenly divided number ending in zero from which the other two factors may be derived. 17--Sqrt(18,000) = 134+ making our two remaining factors two numbers on both sides of 134 both ending in zero. 18--130x140 = 18,200, 120x150 = 18,000. 19--What do you know? Houston, we have a solution! 20--120 + 125 + 150 = 316 = 711 and 120x125x150x316 = 711,000,000. 21--Therefore, the prices of our four items are $1.20, $1.25, $1.50, and $3.16. 22--Again, $1.20 + $1.25 + $1.50 + $3.16 = $7.11 and $1.20($1.25)$1.50($3.16) = $7.11.

The following is a solution by The Mazeman on AOL, many years ago.

> i need to know 4 numbers that when you add them you will get 7.11 and when > you mulitiply the same nubmbers you will get 7.11 also

I find it easier to work with integers, so let's look for four integers that add up to 711, which have a product of 711,000,000.

What are the prime factors of 711,000,000? Right off the bat, we know that 2^6 and 5^6 have to be factors, because

2^6 * 5^6 = 1,000,000

What are the factors of 711? 711 / 9 = 79. 3^2 * 79

The prime factors of 711 million are

2^6 * 5^6 * 3^2 * 79

Since there are six zeroes, we know that the number five has to be a factor of at least three numbers. Why? It can't be a factor of only two, because then 125, 250, 375, 500, 625 would be the only choices for two of the numbers, and there is no such solution. (That's one of the several hints that I will give you).

It can't be a factor of all four numbers. Why?

Because then every number would end in 5 or 0, and there would be no way to have a sum of 711.

Okay, five is a factor of three numbers. What does that tell us about the fourth number? The fourth number has to end in 1 or 6. Why? The first three numbers have to end in five or zero. The only possible end digits are: 5+5+5+6=21 5+5+0+1=11 5+0+0+6=11 0+0+0+1=1

I'm going to give you a hint about the fives. Each of our numbers will have a different power of five as a factor.

One of our numbers will have 5^3 as a factor. It will be a multiple of 125.

One of our numbers will have 5^2 as a factor. It will be a multiple of 25, but not a multiple of 125.

One of our numbers will have 5^1 as a factor. It will be a multiple of 5, but not a multiple of 25.

One of our numbers will have 5^0 as a factor. It will end in 1 or 6, and it must be a multiple of 79. The only multiple of 79 that ends in 6 or 1 is 316.

Are you with me so far?

Let me give you another hint. All of your numbers will be greater than $1 and less than $3.25. In fact, 316 is your biggest number. 711 / 316 = 2.25, so this tells us that the other three numbers must have a product of 2.25.

In other words, our other three numbers must be greater than 100 and less than 225 in whole numbers. The cube root of 2.25 is approximately 1.31, so our three numbers will be relatively close to 1.31.

Here are the multiples of 125 that are greater than 100 and less than 225.

125 = 5^3

Here are the multiples of 25, not shown above, that are in the same range:

150 = 2^1 * 3^1 * 5^2 200 = 2^3 * 5^2

Here are the multiples of 5, not shown above, that are in the same range:

120 = 2^3 * 3^1 * 5^1 180 = 2^2 * 3^2 * 5^1

And we already know that the non-multiple of 5 has to be:

316 = 2^2 * 79^1

I'm going to let you decide from here. You already know that 125 and 316 have to be two of your numbers. What do the other two numbers have to be? Let me know what you figure out.

someguy

Posted - 05/21/2013 : 21:50:18 Hi thehill, my take on where the equations come from is as follows.

If you let a, b, c, and d be the cost in cents, then a/100, b/100, c/100, and d/100 are the costs of each item in dollars.

We are told the sum of the prices equals $7.11. (a/100) + (b/100) + (c/100) + (d/100) = 7.11 Multiplying through by 100 gives a+b+c+d=711.

We are told the product of the prices is $7.11. (a/100)*(b/100)*(c/100)*(d/100)=711/100 Multiplying both sides by (100)^4 to clear the denominator gives abcd=711*(100)^3=711000000

I haven't figured out a way to do it by hand yet, but did write a small bit of code to find the solution. I need to find a way to cut down the number of ways to group the primes together to make the problem small enough to handle by hand.

the_hill1962

Posted - 05/21/2013 : 15:59:45 You need to be looking at factors of 71100000000, not 711 because when you multiply the four items, the "cents" make an 8 place decimal. i.e. "1" means 1 cent === 0.01 .01*.01*.01*.01=.00000001, not 1. Again, the problem got the best of me and I had to look up the solution.

Sure enough, it involves solving a+b+c+d=711 abcd=711000000

Not a+b+c+d=711 abcd=711

Note, I had originally thought 71100000000, not 711000000. I gather that it is 711000000 because the two ones count as part of the 8 place decimal.

Ultraglide

Posted - 05/19/2013 : 18:52:57 Oops, helps if you read the question.

TchrWill

Posted - 05/19/2013 : 10:40:52

quote:Originally posted by Ultraglide

We have to look at the factors of 711 i.e. 3, 3, and 79, so one solution would be 1,3,3,79. Other possibilities are 1,1,9,79 or 1,1,1,711.

Remember that

a + b + c + d = a(b)c(d) = $7.11

Ultraglide

Posted - 05/18/2013 : 16:23:28 We have to look at the factors of 711 i.e. 3, 3, and 79, so one solution would be 1,3,3,79. Other possibilities are 1,1,9,79 or 1,1,1,711.