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Multiplied Prices

by Andrew
(New York)










































A customer brought four items to the cashier of a convenience store. The clerk rang up the items and said, “Okay, I multiplied the prices together and your total cost is $7.11.”
“You multiplied them?” the customer asked. “You’re supposed to add them, you know.”
“I know,” said the clerk, “but it doesn’t make any difference. The total is still $7.11.”
What were the prices of the four items?

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May 06, 2012
Multiplied Prices
by: Staff


Part I

Question:

by Andrew
(New York)

A customer brought four items to the cashier of a convenience store. The clerk rang up the items and said, “Okay, I multiplied the prices together and your total cost is $7.11.”
“You multiplied them?” the customer asked. “You’re supposed to add them, you know.”
“I know,” said the clerk, “but it doesn’t make any difference. The total is still $7.11.”
What were the prices of the four items?


Answer:

Solve the problem using integers (cents rather than dollars and cents)

N₁ + N₂ + N₃ + N₄ = 7.11

N₁ * N₂ * N₃ * N₄ = 7.11

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Solve the problem using integers (cents, rather than dollars and cents)


N₁ + N₂ + N₃ + N₄ = 7.11


100N₁ + 100N₂ + 100N₃ + 100N₄ = 100*7.11

100N₁ + 100N₂ + 100N₃ + 100N₄ = 711



N₁ * N₂ * N₃ * N₄ = 7.11

100N₁ * 100N₂ * 100N₃ * 100N₄ = 10⁸*7.11

100N₁ * 100N₂ * 100N₃ * 100N₄ = 711,000,000

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The two equations:

100N₁ + 100N₂ + 100N₃ + 100N₄ = 711

100N₁ * 100N₂ * 100N₃ * 100N₄ = 711,000,000


Or

a + b + c + d = 711

a * b * c * d = 711,000,000


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711000000 = 2⁶ * 3² * 5⁶ * (1*79)


One of the numbers (“a”) must be a multiple of 79, which is a factor of 711000000.

This reduces the choices for one number to: 79 (x1), 158 (x2), 237 (x3), 316 (x4), 395 (x5), 474 (x6), 553 (x7), 632 (x8), 711 (x9)

When multiplying b*c*d, the maximum product occurs when all three numbers are equal (b=c=d). Because of this restriction, the product of b*v*c occurs when [(b+c+d)/3]³, or (b+c+d)³/27. This rules out a =395, a = 474, a= 553, a = 632, and a = 711. Calculations are shown below.


79 * 1 = 79
If a = 79, then b+c+d = 632, and b*c*d = 9000000 (which is 711000000/79)
711000000 = 2⁶ * 3² * 5⁶ * (1*79)
632^3/27 = 9349480.2962962966
9349480 > the required 9000000 – a = 79, possible solution





79 * 2 = 158
If a = 158, then b+c+d = 553, and b*c*d = 4500000 (which is 711000000/158)
711000000 = 2⁵ * 3² * 5⁶ * (2*79)
553^3/27 = 6263421.3703703703
6263421 > the required 4500000 – a = 158, possible solution


79 * 3 = 237
If a = 237, then b +c+d=474, and b*c*d = 3000000 (which is 711000000/237)
711000000 = 2⁶ * 3 * 5⁶ * (3*79)
474^3/27 = 3944312
3944312> the required 3000000 – a = 237, possible solution


---------------------------------------------------------------


May 06, 2012
Multiplied Prices
by: Staff


---------------------------------------------------------------

Part II


79 * 4 = 316
If a = 316, then b+c+d=395, and b*c*d = 2250000 (which is 711000000/316)
711000000 = 2⁴ * 3² * 5⁶ * (2²*79)
395^3/27 = 2282587.9629629632
2282587 > the required 2250000 – a = 316, possible solution



79 * 5 = 395
If a = 395, then b+c+d=316, and b*c*d = 1800000 (which is 711000000/395)
316^3/27 = 1168685.0370370371
1168685 < the required 1800000 – not a possible solution

79 * 6 = 474
If a = 474, then b+c+d=237, and b*c*d = 1500000 (which is 711000000/474)
237^3/27 = 493039
493039 < the required 1500000 – not a possible solution

79 * 7 = 553
If a = 553, then b+c+d=158, and b*c*d = 1285714.2857142857 (which is 711000000/553)
158^3/27 = 146085.62962962964
146085 < the required 1285714 – not a possible solution

79 * 8 = 632
If a = 632, then b+c+d=79, and b*c*d = 1125000 (which is 711000000/632)
79^3/27 = 18260.703703703704
18260 < the required 1125000 – not a possible solution


79 * 9 = 711– not a possible solution
If a = 711, then b+c+d=0, and b*c*d = 1000000 (which is 711000000/711)
0 < the required 1000000 – not a possible solution

------------------------------

711000000 = 2⁶ * 3² * 5⁶ * (1*79)

“a” has only four possible values: 79, 158, 237, 316


2⁶= 64,which could be a factor of only one number.

3² = 9, which could be the factor on only one number.

5⁶ = 15625. It is too large to be the factor of one number. A number is limited to three digits.

5⁴ = 625. It is also too large to be the factor of one number because it is too close to 711.

5³ = 125, a possibility.

AT LEAST two numbers must be multiples of 5 (5³ * 5³ = 5⁶).



Therefore at least one of the numbers must be a multiple of 2, at least one of the numbers must be a multiple of 3, and at least two of the numbers must be a multiple of 5.



Since there are only 4 possible choices for a (79, 158, 237, 316), you can solve for possible values of d (using the idea that two of the numbers must be multiples of 5).

If a = 79, then b+c+d = 632
d = 632-125-125 = 382
d = 632-2*125-125 = 257
d = 632-3*125-125 = 132
d = 632-4*125-125 = 7


If a = 158, then b+c+d = 553
d = 553-125-125 = 303
d = 553-2*125-125 = 178
d = 553-3*125-125 = 53


If a = 237, then b +c+d=474
d = 474-125-125 = 224
d = 474-2*125-125 = 99


If a = 316, then b+c+d=395
d = 395-125-125 = 145
d = 395-2*125-125 = 20


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May 06, 2012
Multiplied Prices
by: Staff


---------------------------------------------------------------

Part III


d (possible values) = 382, 257, 132, 7, 303, 178, 53, 224, 99, 145, 20


d (possible value factored) = 382 = 2*191 – eliminated – prime factor different than 2, 3, or 5

d (possible value factored) = 257 = 257 (prime) – eliminated – prime factor different than 2, 3, or 5
d (possible value factored) = 132 = 2²*3*11 – eliminated – prime factor different than 2, 3, or 5

d (possible value factored) = 7 = 7 (prime) – prime factor different than 2, 3, or 5

d (possible value factored) = 303 = 3*101 – eliminated – prime factor different than 2, 3, or 5

d (possible value factored) = 178 = 2*89 – eliminated – prime factor different than 2, 3, or 5

d (possible value factored) = 53 = 53 (prime) – prime factor different than 2, 3, or 5

d (possible value factored) = 224 = 2⁵*7 – eliminated – prime factor different than 2, 3, or 5

d (possible value factored) = 99 = 3²*11 – eliminated – prime factor different than 2, 3, or 5

d (possible value factored) = 145 = 5*29 – eliminated – prime factor different than 2, 3, or 5

d (possible value factored) = 20 = 2²*5





the ONLY value of d (d = 20) which works occurs when “a” = 316

Therefore, “a” does = 316

79 * 4 = 316
a = 316, b+c+d =395, and b*c*d = 2250000 (which is 711000000/316)
711000000 = 2⁴ * 3² * 5⁶ * (2²*79)


From this point on, trial and error seems to work best.

The final answer is:

N₁ = $3.16
N₂ = $1.50
N₃ = $1.25
N₄ = $1.20




Thanks for writing.

Staff
www.solving-math-problems.com





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