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Consider two integers: an original two digits integer and the integer

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Consider two integers: an original two digits integer and the integer  [#permalink]

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New post Updated on: 17 Apr 2018, 14:38
00:00
A
B
C
D
E

Difficulty:

  75% (hard)

Question Stats:

50% (01:23) correct 50% (01:28) wrong based on 22 sessions

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Consider two integers: an original two digits integer and the integer formed by reversing the digits of this original integer. If the bigger of these two integers is divided by the other integer, what could be the maximum remainder?
1) 9
2)27
3)36
4)45
5) 54

Originally posted by Nina1987 on 17 Apr 2018, 05:49.
Last edited by Nina1987 on 17 Apr 2018, 14:38, edited 1 time in total.
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Re: Consider two integers: an original two digits integer and the integer  [#permalink]

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New post 17 Apr 2018, 06:28
32/23 gives remainder 9.

42/24 gives remainder 18.

94/49 gives remainder 45.

98/89 gives remainder 9.

Likewise, by putting certain values, answer can be determined. Thus, maximum is 45.

Hence option D.

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Re: Consider two integers: an original two digits integer and the integer  [#permalink]

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New post 17 Apr 2018, 06:34
1
SonalSinha803 wrote:
32/23 gives remainder 9.

42/24 gives remainder 18.

94/49 gives remainder 45.

98/89 gives remainder 9.

Likewise, by putting certain values, answer can be determined. Thus, maximum is 45.

Hence option D.

Posted from my mobile device


HOw did you choose those numbers? Randomly or is there any method?
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Re: Consider two integers: an original two digits integer and the integer  [#permalink]

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New post 17 Apr 2018, 08:23
Nina1987 wrote:
SonalSinha803 wrote:
32/23 gives remainder 9.

42/24 gives remainder 18.

94/49 gives remainder 45.

98/89 gives remainder 9.

Likewise, by putting certain values, answer can be determined. Thus, maximum is 45.

Hence option D.

Posted from my mobile device


HOw did you choose those numbers? Randomly or is there any method?



I am also confused as to how you chose those numbers to test with.
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Re: Consider two integers: an original two digits integer and the integer  [#permalink]

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New post 17 Apr 2018, 08:52
Nina1987 wrote:
If a two digit integer is divided by an integer formed by reversing its digits, what could be the maximum remainder?

A. 9
B. 27
C. 36
D. 45
E. 54


How about 89 and 98 when 89 in nominator and 98 in denominator? 89>45

I think that the question is incorrect.
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Re: Consider two integers: an original two digits integer and the integer  [#permalink]

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New post 17 Apr 2018, 14:39
Hero8888 wrote:
Nina1987 wrote:
If a two digit integer is divided by an integer formed by reversing its digits, what could be the maximum remainder?

A. 9
B. 27
C. 36
D. 45
E. 54


How about 89 and 98 when 89 in nominator and 98 in denominator? 89>45

I think that the question is incorrect.


sorry just edited the question. pls check now.
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Re: Consider two integers: an original two digits integer and the integer  [#permalink]

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New post 17 Apr 2018, 20:28
1
Since this is a max problem, one strategy is to see if the max answer choice (Option E) fits. If it does, you've found the answer. If not, try Option D.

To maximize the remainder, we need to maximize the smaller 2-digit integer. The smaller 2-digit integer is the divisor, and the remainder is less than the divisor. So if the divisor is 12, most we can hope for is a remainder of 11. On the other hand, if the divisor is 60, we can have a remainder as high as 59. But note that the max remainder we can have when a larger 2-digit integer is divided by a smaller 2-digit integer is 49 (99/50). This eliminates Option E.

Let's see if Option D can work. I'll leave it to you to see why having a denominator greater than 50 wont give you a remainder of 45. Let's go the other way and try the next denominator down, 49 --> 94/49 yields a remainder of 45, which is Option D.

Answer: D
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Re: Consider two integers: an original two digits integer and the integer &nbs [#permalink] 17 Apr 2018, 20:28
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Consider two integers: an original two digits integer and the integer

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