 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Sum Number and Reversal (Posted on 2016-06-24) It is observed that the sum of 2016 and its reversal 6102 is 8118 and, 8118 is divisible by 41.

Determine the five positive integers following 2016 with the property that the sum of the number and its reversal is divisible by 41.

*** As an extra challenge, try to solve this problem without using a computer program assisted method.

 No Solution Yet Submitted by K Sengupta No Rating Comments: ( Back to comment list | You must be logged in to post comments.) No Subject | Comment 2 of 5 | 2016 is 7 (mod 41). So 6102 is 34 (mod 41)
2017 is 8 (mod 41). 7102 is 9 (mod 41)
2018 is 9 (mod 41). 8102 is 25 (mod 41)
2019 is 10 (mod 41). 9012 is 41=0 (mod 41)
The sum of both congruencies in each line is the congruence of the sum of both numbers.
It gives a simple sequence: 41=0, 17, 34, 51=10, ...

2020 is 11 (mod 41) 0202 is 38 (mod 41). The sum is 49=8 instead of 68=27 as expected. So there is a jump.

This jump is going to happen for each ten numbers, when the reversed number changes from 9xxx to 0xxx.

In the interval (2016-2099) there is only this type of jump. In it the sequence not only do not adds 17 units but also goes back 2 units.

So for this interval it is possible to calculate the congruence to mod 41 of the sum of the pair of numbers through the formula:
(17*(n+3)-19m) (mod 41). Where n=distance from origin +3 (=2019) and m=ten digit of n plus 1. It gives the sequence:
0, 17, 34, 51=10, 49=8, 25, 42=1, 18, 35, ...

When the sequence takes the value 41=0 there will be a compliant number. It happens for n=46 (distance from 2016=49). So for 2065.

2065+5602=7667=41*187

As the sequence is linear it could be expected that after other 49 numbers we should find another occurrence, but it is not the case because number 2100 introduce another jump.

Using the formula it should be 17*83-19*9=1259=29 But instead it is 21 (2100+0012=2112 is 21 mod 41)

In the interval 2100-2199 the formula changes to 17(n+3)-19m-6, so that no 49 numbers ahead but 41 we will have another occurrence.
2106+6012=8118 (same sum for 2016)
And 49 ahead another 2155. But next occurs again the hundred digits jump...

So 2016, 2065, 2106, 2155,

Edited on June 25, 2016, 6:37 pm

Edited on June 26, 2016, 6:48 am
 Posted by armando on 2016-06-25 18:09:34 Please log in:

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