I was given the following question for homework and I cannot figure it out! please help!! its due at 8 am tomorrow!!!
A hurried shopper in a hardware store used a calculator to add up his purchases. Carelessly, he presses x each times instead of +. Amazingly, after the first three purchases, the calculator shows the correct sum. When he enters the price of the fourth item and presses x again, the calculator shows the correct sum of all four items. All prices are different and no fractions of a cent are used. What are the four prices? (Hint: A perfect number is involved.)
Based on the wise words of galactus, I offer you the following.
<< 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? >>
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.
.........................................................SUM
x......79x.........N.............N^(1/3).....N + 3[N^(1/3)]
2.....158....4,500,000........165+............653
3.....237....3,000,000........144+............669
4.....316....2,250,000........131+............709
5.....395....1,800,000........121+............758
6.....474....1,500,000........114+............816
7.....553....1,285,714+......108+............877
8.....632....1,125,000........104..............944
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:
.N.......2,250,000/N
.95........23,684+
105.......21,428+
115.......19,565+
125.......18,000
135.......16,666+
145.......15,517+
155.......14,516+
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.
There are many other similar $ amounts that have similar solutions. You can find 160 of them at
The Internet Center for Mathematics Problems
