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The "reflection" of a positive integer is obtained by reversing its digits. For example 321 is the reflection of 123 The difference between a five-digit integer and its reflection must be divisible by which of the following?
A. 2
B. 4
C. 5
D. 6
E. 9
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A. 2
B. 4
C. 5
D. 6
E. 9
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Number = abcde = 10^4 ( a ) + 10^3 (b) + 10^2 (c) + 10 (d) + e
Reflection = edcba = 10^4 ( e ) + 10^3 (d) + 10^2 (c) + 10 (b) + a
Number - Reflection = 10^4 ( a - e ) + 10^3 ( b -d ) + 10 (d- b) + e - a
= (a-e) ( 10 ^ 4 - 1) + (b-d) ( 10^3 - 10)
= (a-e) (99999) + (b-d) (990)
= 9* (something)
Hence the answer.
Reflection = edcba = 10^4 ( e ) + 10^3 (d) + 10^2 (c) + 10 (b) + a
Number - Reflection = 10^4 ( a - e ) + 10^3 ( b -d ) + 10 (d- b) + e - a
= (a-e) ( 10 ^ 4 - 1) + (b-d) ( 10^3 - 10)
= (a-e) (99999) + (b-d) (990)
= 9* (something)
Hence the answer.
General Discussion
Archit3110
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Bunuel
eg = 12345 and reflection 54321 , 45789 , reflection 98754
testing the values , difference would always be divisible by 9
IMO E
KanishkM
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Bunuel
when the number is taken as 54321, reflection will be 12345, difference will be / by 2 4 and 9
when the number is taken as 76432, reflection will be 23467, difference will be / 9
Answer E
antartican
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Bunuel
What is the mathematical concept behind this ?
I, like everyone else, did the examples.
Posts: 8
The difference should be always a even number, and as it is divided by 9 so it should be divided by 6 as well, so out of "2", "6" and "9", why are we always taking "9" only?
**Need Explanation
**Need Explanation
Posts: 1,052
Updated on: 23 Jun 2020, 11:38
Originally posted by rajatchopra1994 on 23 Jun 2020, 11:22.
Last edited by rajatchopra1994 on 23 Jun 2020, 11:38, edited 1 time in total.
Last edited by rajatchopra1994 on 23 Jun 2020, 11:38, edited 1 time in total.
indugopal1991
Dear indugopal1991 ,
Take example :1. 12345 its reflection is 54321 is divisible2,6, 9.
2. 45789 its reflection 98754 is only divisible 9
3. 34567 its reflection 76543
For all 5 digit number reflection, it is only divisible by 9. Sometime may or may not be it is divisible by any other number apart from 9
Regards,
Rajat Chopra
Posts: 8
Dear Raj, for the first example, 54321-12345=41976, which is divisible by 2, 6 and 9.
But for the second value 45679 it will not work.
But for the second value 45679 it will not work.
antartican
It is like this:
Let the 5 digit number look like ABCDE, where each letter signifies the digit in each of the 5 places
So its reflection is EDCBA
Value of ABCDE = \(10^4A + 10^3B + 10^2C + 10D + E\)
Value of EDCBA = \(10^4E + 10^3D + 10^2C + 10B + A\)
Difference = \((10^4-1)A + (10^3-10)B + (10-10^3)D + (1-10^4)E\)
= \(9999A + 990B - 990D - 9999E \)
= \(9(1111A + 110B - 110D - 1111E)\)
So it is divisible by 9
Answer: E
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Expert reply
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Bunuel
Solution:
We can let 12,345 be the 5-digit integer, so its reflection is 54,321 and their difference is 54,321 - 12,345 = 41,976. At this point, we see that we can only eliminate choice C as the correct answer.
Let’s use another 5-digit integer, say 12,344. Its reflection is 44,321 and their difference is 44,321 - 12,344 = 31,977. We see that this number can be divisible only by 9, so choice E must be the correct answer.
Answer: E
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Let no.: abcde, and its mirror is edcba
abcde - edcba:
If b>d [(e+10-a) (d-1+10-b) (c-1+10-c) (b-1-d) (a-e)]
Sum equals to 27 (divisible by 9)
If b<d [(e+10-a) (d-1-b) (c-c) (b+10-d) (a-1-e)
Sum equals to 18 (divisible by 9)
Hence must be multiplied by 9
Posted from my mobile device
abcde - edcba:
If b>d [(e+10-a) (d-1+10-b) (c-1+10-c) (b-1-d) (a-e)]
Sum equals to 27 (divisible by 9)
If b<d [(e+10-a) (d-1-b) (c-c) (b+10-d) (a-1-e)
Sum equals to 18 (divisible by 9)
Hence must be multiplied by 9
Posted from my mobile device
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