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Re: How many three-digit integers exist such that all their digi [#permalink]

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24 Dec 2011, 04:49

The hundreds digit can be filled in 4 ways (2,4,6,8 but not 0 as in that case the number will become two digit instead of three digit) and the next two can be filled in 5 ways each. Therefore total such numbers = 4 x 5 x 5 = 100
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Re: How many three-digit integers exist such that all their digi [#permalink]

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24 Dec 2011, 10:02

GyanOne wrote:

The hundreds digit can be filled in 4 ways (2,4,6,8 but not 0 as in that case the number will become two digit instead of three digit) and the next two can be filled in 5 ways each. Therefore total such numbers = 4 x 5 x 5 = 100

The question says : All digits must be even. Is zero considered even on GMAT? I know the approach, but this is my doubt. If not, answer should be 4*4*4 = 64.
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Re: How many three-digit integers exist such that all their digi [#permalink]

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24 Dec 2011, 19:37

vailad wrote:

GyanOne wrote:

The hundreds digit can be filled in 4 ways (2,4,6,8 but not 0 as in that case the number will become two digit instead of three digit) and the next two can be filled in 5 ways each. Therefore total such numbers = 4 x 5 x 5 = 100

The question says : All digits must be even. Is zero considered even on GMAT? I know the approach, but this is my doubt. If not, answer should be 4*4*4 = 64.

In Gmat 0 is considered Even so 100 is the answer
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Re: How many three-digit integers exist such that all their digi [#permalink]

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10 Jan 2013, 19:06

dreamchase wrote:

How many three-digit integers exist such that all their digits are even?

80 100 120 125 135

I happened to solve it in less than 30 seconds but wouldn't suggest the approach. Based on the question, range in which the numbers suggested can exist is from 200 - 888 (both inclusive). It all starts as 200, 202,204,206,208 --> 5 numbers. Again it will should start from 220,222,....228 ----> 5 numbers

In 200's, we will have 25 numbers like that. Again, 300's cant have any, nor can 500's or 700's. Thus, we will be getting valid numbers only from 200's, 400's, 600's, 800's ----> 25*4=100.

If you can think fast, then you can narrow it on to the answer, sooner.
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Re: How many three-digit integers exist such that all their digi [#permalink]

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12 Jan 2013, 23:28

anuj4ufriends wrote:

I happened to solve it in less than 30 seconds but wouldn't suggest the approach. Based on the question, range in which the numbers suggested can exist is from 200 - 888 (both inclusive). It all starts as 200, 202,204,206,208 --> 5 numbers. Again it will should start from 220,222,....228 ----> 5 numbers

In 200's, we will have 25 numbers like that. Again, 300's cant have any, nor can 500's or 700's. Thus, we will be getting valid numbers only from 200's, 400's, 600's, 800's ----> 25*4=100.

If you can think fast, then you can narrow it on to the answer, sooner.

Well, I still did it in less than 10 seconds using the 4*5*5 approach. I was recently helping out a friend to prepare for his GMAT, and he was amazed by the simplicity of this approach in a very similar problem. He was also trying to count the numbers, but he was not nearly as fast... It reminds me the story about John von Neumann and the Two Trains puzzle. In case you don't remember, here is the puzzle:

Two trains are on the same track a distance 100 km apart heading towards one another, each at a speed of 50 km/h. A fly starting out at the front of one train, flies towards the other at a speed of 75 km/h. Upon reaching the other train, the fly turns around and continues towards the first train. How many kilometers does the fly travel before getting squashed in the collision of the two trains?

John von Neumann is reputed to have immediately answered with the correct result. When subsequently asked if he had heard the short-cut solution, he answered no, that his immediate answer had been a result of explicitly summing the series.

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Re: How many three-digit integers exist such that all their digi [#permalink]

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12 Jan 2013, 23:31

SergeyOrshanskiy wrote:

anuj4ufriends wrote:

I happened to solve it in less than 30 seconds but wouldn't suggest the approach. Based on the question, range in which the numbers suggested can exist is from 200 - 888 (both inclusive). It all starts as 200, 202,204,206,208 --> 5 numbers. Again it will should start from 220,222,....228 ----> 5 numbers

In 200's, we will have 25 numbers like that. Again, 300's cant have any, nor can 500's or 700's. Thus, we will be getting valid numbers only from 200's, 400's, 600's, 800's ----> 25*4=100.

If you can think fast, then you can narrow it on to the answer, sooner.

Well, I still did it in less than 10 seconds using the 4*5*5 approach. I was recently helping out a friend to prepare for his GMAT, and he was amazed by the simplicity of this approach in a very similar problem. He was also trying to count the numbers, but he was not nearly as fast... It reminds me the story about John von Neumann and the Two Trains puzzle. In case you don't remember, here is the puzzle:

Two trains are on the same track a distance 100 km apart heading towards one another, each at a speed of 50 km/h. A fly starting out at the front of one train, flies towards the other at a speed of 75 km/h. Upon reaching the other train, the fly turns around and continues towards the first train. How many kilometers does the fly travel before getting squashed in the collision of the two trains?

John von Neumann is reputed to have immediately answered with the correct result. When subsequently asked if he had heard the short-cut solution, he answered no, that his immediate answer had been a result of explicitly summing the series.

That's the reason why I said I dont suggest it.
_________________

Re: How many three-digit integers exist such that all their digi [#permalink]

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19 Jun 2014, 02:46

Bunuel,

This is Quantitative :: Problem solving :: Probability & Combinations :: M13-14

0 is also an even integer so last digit can be filled in 5 ways- 0,2,4,6,8

for both second and third digit there are 5 possibilities-

5 X 5 X 5 = 125 _______________

This is the solution given at Gmatclubtest

The first digit can be any of the four: 2, 4, 6, or 8. For both second and third digits, there are 5 possibilities. The answer is 4∗5∗5=100. The correct answer is B
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How many three-digit integers exist such that all their digits are even?

A. 80 B. 100 C. 120 D. 125 E. 135

Bunuel,

This is Quantitative :: Problem solving :: Probability & Combinations :: M13-14

0 is also an even integer so last digit can be filled in 5 ways- 0,2,4,6,8

for both second and third digit there are 5 possibilities-

5 X 5 X 5 = 125 _______________

This is the solution given at Gmatclubtest

The first digit can be any of the four: 2, 4, 6, or 8. For both second and third digits, there are 5 possibilities. The answer is 4∗5∗5=100. The correct answer is B

If the first digit is 0, then the number becomes two-digit, not three-digit.
_________________

Re: How many three-digit integers exist such that all their digi [#permalink]

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19 Jun 2014, 02:52

Bunuel wrote:

honchos wrote:

How many three-digit integers exist such that all their digits are even?

A. 80 B. 100 C. 120 D. 125 E. 135

Bunuel,

This is Quantitative :: Problem solving :: Probability & Combinations :: M13-14

0 is also an even integer so last digit can be filled in 5 ways- 0,2,4,6,8

for both second and third digit there are 5 possibilities-

5 X 5 X 5 = 125 _______________

This is the solution given at Gmatclubtest

The first digit can be any of the four: 2, 4, 6, or 8. For both second and third digits, there are 5 possibilities. The answer is 4∗5∗5=100. The correct answer is B

If the first digit is 0, then the number becomes two-digit, not three-digit.

My fault, i thought with first digit you mean unit digit. I took the explanation in the reverse order. My bad I am sorry.

Solution was slightly confusing, if terms like unit,tens and 100th digits were used the confusion would have been avoided.

Thanks Bunuel.
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