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Hi What is the logic for dividing by 5 or 10 factorial?
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Hi What is the logic for dividing by 5 or 10 factorial?

Where do we divide by 5! or 10! ?
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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

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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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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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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liranmaymoni
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

how can W be 4 . the product of weight is given . now when we are left with 625 which is 5W, where W is the no. of widgets.

we need t divide it by 5 to get the no. of widgets.

Regards,
Kumar Gaurav
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Seems as though the easiest 1st step to take is to look for the Number of Trailing Zeroes this large number has.

10, 435, 040, 000 = (1, 043, 504) * 10,000

Since 1, 043, 504 is not a Multiple of 5, we can ignore that part of the Product.

10, 000 = (10)^4th = (2)^4 * (5)^4


There are thus FOUR Prime Bases of 5 that compose the number.

When all the individual weights are multiplied, there must only be 4 Widgets weighing 5 lbs. each

Answer -C-

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liranmaymoni
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

Similarly, would d be four as well with the same approach?
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liranmaymoni
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
Solution:

We need to prime factorize 10,435,040,000:

10,435,040,000 = 260,876 x 40,000 = 14^2 x 11^3 x 2^2 x 10,000 = 2^2 x 7^2 x 11^3 x 2^2 x 2^4 x 5^4 = 2^8 x 11^3 x 5^4 x 7^2

Since each widget weighs 5 pounds, there are 4 widgets.

Alternate Solution:

Since each widget weighs 5 pounds, we are interested only in how many times 5 divides into 10,435,040,000. We can express this number as 1,043,504 x 10,000 = 1,043,504 x 5^4 x 2^4. Since 5 cannot further divide into 1,043,504, we see that 4 is the greatest power of 5 in the factorization of 10,435,040,000. Thus, we see that the cargo ship unloaded 4 widgets.

Answer: C
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We're told the product of the individual weights of the unloaded items equal 10,435,040,000 pounds. The word product is a hint that we need the prime factorization.

10,435,040,000

= 1,043,504 * 10^4
= 1,043,504 * 5^4 * 2^4

We're looking for the number of widgets, which weighs 5 pounds each. In our prime factorization, the five is raised to four. Therefore, we have 4 widgets. Answer is C.
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liranmaymoni
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

Similarly, would d be four as well with the same approach?

The same method can be used to determine the values of d, g and y; however, the value of d is not 4, it is 8. The number 10,435,040,000 can be prime factorized as 2^8 × 5^4 × 7^2 × 11^3 and d is the exponent of 2; hence, d = 8.
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If you recognize that the power of 5 is what's required and that 5^4 is 625, then it seems likely that 625 is the trick answer and that therefore C is likely the correct answer
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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?

\(10,435,040,000 = 2^8*11^3*5^4*7^2\)

Let the number of doohickies, geegaws, widgets, and yamyams be d, g, w & y respectively

The product of the individual weights of the unloaded items = \(2^d*11^g*5^w*7^y = 10,435,040,000 = 2^8*11^3*5^4*7^2\)
d = 8; g = 3; w = 4; y = 2

The number of widgets unloaded = w = 4

IMO C
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