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A cargo ship carrying four kinds of items, doohickies, geega

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A cargo ship carrying four kinds of items, doohickies, geega [#permalink]

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A cargo ship carrying four kinds of items, doohickies, geegaws, widgets, and yamyams, arrives at the port. Each item weighs 2, 11, 5, and 7 pounds, respectively, and each item is weighed as it is unloaded. If, in the middle of the unloading process, the product of the individual weights of the unloaded items equals 10,435,040,000 pounds, how many widgets have been unloaded?

A. 2
B. 3
C. 4
D. 625
E. 2,087,008,000

The OA say we should break the number into primes.. "i'm not sure how to do it with a number this big without taking FOREVER

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Re: A cargo ship carrying four kinds of items, doohickies, geega [#permalink]

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New post 15 Feb 2014, 03:36
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Hi liranmaymoni,

you need to know the number of widgets which weight 5 pounds each. This is a very common type of exercise where basically you have a number and need to know how many times a certain prime divides the number.

Here one way to solve the exercise would be to find the full decomposition of the number. This will probably take some time but, however, there is a way to do way faster. You don't need the full decomposition, what you need is just the number of times "5" divides the given huge number which is closely related to the number of times 10 divides the number. Luckily it is really easy to divide a number by 10 !

10,435,040,000= 10,000 * 10,435,04 = 10^4 * 10,435,04 = 5^4 * 2^4 * 10,435,04

Since the last number cannot be divided by 5 (because it does not end either by 0 or 5) you know that the initial number can be divided exactly 4 times by 5 and here comes your answer!
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Re: A cargo ship carrying four kinds of items, doohickies, geega [#permalink]

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New post 15 Feb 2014, 03:40
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liranmaymoni wrote:
A cargo ship carrying four kinds of items, doohickies, geegaws, widgets, and yamyams, arrives at the port. Each item weighs 2, 11, 5, and 7 pounds, respectively, and each item is weighed as it is unloaded. If, in the middle of the unloading process, the product of the individual weights of the unloaded items equals 10,435,040,000 pounds, how many widgets have been unloaded?

A. 2
B. 3
C. 4
D. 625
E. 2,087,008,000

The OA say we should break the number into primes.. "i'm not sure how to do it with a number this big without taking FOREVER


The only thing you are interested is the power of 5 in 10,435,040,000:

\(10,435,040,000=1,043,504*10,000=1,043,504*10,000=1,043,504*(2^4*5^4)\).

Answer: C.

Similar questions to practice:
in-a-certain-game-a-large-container-is-filled-with-red-yel-144902.html (OG PS)
in-a-certain-game-a-large-bag-is-filled-with-blue-green-126425.html (MGMAT PS)

Hope it helps.
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Re: A cargo ship carrying four kinds of items, doohickies, geega [#permalink]

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New post 24 Oct 2014, 23:39
Hi What is the logic for dividing by 5 or 10 factorial?
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Re: A cargo ship carrying four kinds of items, doohickies, geega [#permalink]

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Re: A cargo ship carrying four kinds of items, doohickies, geega [#permalink]

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New post 15 Dec 2014, 09:50
Bunuel wrote:
liranmaymoni wrote:
A cargo ship carrying four kinds of items, doohickies, geegaws, widgets, and yamyams, arrives at the port. Each item weighs 2, 11, 5, and 7 pounds, respectively, and each item is weighed as it is unloaded. If, in the middle of the unloading process, the product of the individual weights of the unloaded items equals 10,435,040,000 pounds, how many widgets have been unloaded?

A. 2
B. 3
C. 4
D. 625
E. 2,087,008,000

The OA say we should break the number into primes.. "i'm not sure how to do it with a number this big without taking FOREVER


The only thing you are interested is the power of 5 in 10,435,040,000:

\(10,435,040,000=1,043,504*10,000=1,043,504*10,000=1,043,504*(2^4*5^4)\).

Answer: C.

Similar questions to practice:
in-a-certain-game-a-large-container-is-filled-with-red-yel-144902.html (OG PS)
in-a-certain-game-a-large-bag-is-filled-with-blue-green-126425.html (MGMAT PS)

Hope it helps.


That's a really neat way to do it. I just did division by 5 over and over, but it's way faster to do it this way. Thanks for the tip!
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Re: A cargo ship carrying four kinds of items, doohickies, geega [#permalink]

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New post 15 Dec 2014, 21:28
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jwang27 wrote:

That's a really neat way to do it. I just did division by 5 over and over, but it's way faster to do it this way. Thanks for the tip!


Remember that any number divisible by 5 will always end in a 0 or a 5.
So if you have a number with non zero digits and then 0s at the end, it can easily be split up into the non zero component and the 0s component:

304260000 = 30426 * 10000
30426 is certainly not divisible by 5. You just have to focus on 10000

Mind you, be careful if the non zero digit is 5:
110075000 = 110075 * 1000
Here, 110075 will also be divisible by 5 so you will need to factorize it too.
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Re: A cargo ship carrying four kinds of items, doohickies, geega [#permalink]

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New post 30 Apr 2018, 16:38
liranmaymoni wrote:
A cargo ship carrying four kinds of items, doohickies, geegaws, widgets, and yamyams, arrives at the port. Each item weighs 2, 11, 5, and 7 pounds, respectively, and each item is weighed as it is unloaded. If, in the middle of the unloading process, the product of the individual weights of the unloaded items equals 10,435,040,000 pounds, how many widgets have been unloaded?

A. 2
B. 3
C. 4
D. 625
E. 2,087,008,000



We can let d, g, w, and y be the number of doohickies, geegaws, widgets, and yamyams respectively and create the following equation:

2^d x 11^g x 5^w x 7^y = 10,435,040,000

Notice that we are asked only for the number of widgets, or the value of w. We observe that 10,435,040,000 has 4 trailing zeros, which means there must be 4 factors of 5 in 10,435,040,000. Thus w must be 4.

Answer: C
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Re: A cargo ship carrying four kinds of items, doohickies, geega   [#permalink] 30 Apr 2018, 16:38
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