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22 Oct 2003, 20:51
Kudos
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Which of the following is(are) a multiple of 54 x 143 x 65
I. 52 x 142 x 64
II. 35 x 212 x 512
III. 244 x 253
IV. 154 x 28
How do you do this?
I. 52 x 142 x 64
II. 35 x 212 x 512
III. 244 x 253
IV. 154 x 28
How do you do this?
Archived Topic
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This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Quantitative Questions Forum
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23 Oct 2003, 07:03
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just multiply the unit's digits and check it with each given choice...
the unit's digit of 54 x 143 x 65 is 0.(4 x 3 = 12 => 2 , 2 x 5 = 10=>0)
now check it with each choice..
I. 2 * 2 * 4 = 6
II. 5 * 5 * 2 = 0
III.4 * 3 = 2
IV. 4 * 8 = 2
obviously II is the answer...
if you have any other easier way... please let me know
HTH
Adhi
the unit's digit of 54 x 143 x 65 is 0.(4 x 3 = 12 => 2 , 2 x 5 = 10=>0)
now check it with each choice..
I. 2 * 2 * 4 = 6
II. 5 * 5 * 2 = 0
III.4 * 3 = 2
IV. 4 * 8 = 2
obviously II is the answer...
if you have any other easier way... please let me know
HTH
Adhi
23 Oct 2003, 15:06
Kudos
Bookmarks
I do not think that any of the answer is correct.
Adhi,
The logic you have applied can not be true. Just because the unit digit is 0 it can not be called multiple of another number with 0 unit digit.
For exampple 90 is not multiple of 80. Also, if you use calculator, you will find that II is not the multiple of the given product.
Having said that, I think the problem should be solved in the following way. All the products will need to be displayed in terms of prime numbers.
Answer choice I and IV can not be the answer because it is obvisous that the value of the product in these choices is less than the original product. In order for the answer choice to be multiple of the given product, it has to be greater than the given product. Eliminate I and IV
Answer choice II = 35 x 212 x 512 = 5 X 7 X 2^2 X 53 X 2^9
= 5 X 7 X 53 X 2^11
Original product = 54 x 143 x 65 = 3^3 x 2 x 143 x 5 x 13
In order for this answer choice to be multiple of the given product the divison of the answer choice by the original product should give an integer. Which is not the case here.
Similarly, the answer choice III can be examined but even that choice is not the answer.
Is there anything wrong with the question?
Adhi,
The logic you have applied can not be true. Just because the unit digit is 0 it can not be called multiple of another number with 0 unit digit.
For exampple 90 is not multiple of 80. Also, if you use calculator, you will find that II is not the multiple of the given product.
Having said that, I think the problem should be solved in the following way. All the products will need to be displayed in terms of prime numbers.
Answer choice I and IV can not be the answer because it is obvisous that the value of the product in these choices is less than the original product. In order for the answer choice to be multiple of the given product, it has to be greater than the given product. Eliminate I and IV
Answer choice II = 35 x 212 x 512 = 5 X 7 X 2^2 X 53 X 2^9
= 5 X 7 X 53 X 2^11
Original product = 54 x 143 x 65 = 3^3 x 2 x 143 x 5 x 13
In order for this answer choice to be multiple of the given product the divison of the answer choice by the original product should give an integer. Which is not the case here.
Similarly, the answer choice III can be examined but even that choice is not the answer.
Is there anything wrong with the question?
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Quantitative Questions Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
