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.
There is a better way to approach this puzzle
N (abcd) and M (dcba).
N=1000*a+100*b+10*c+d
M=1000*d+100*c+10*b+a
N+M=1001*(a+d) + 110*(b+c)
This expression should be congruent to 0 (mod 41)
1001 is congruent to 17 (mod 41)
110 is congruent to 28 (mod 41)
So 17*(a+d) + 28*(c+b) congruent to 0 (mod 41)
a+d and c+b can assume values from 0 to 18 as they are digits.
Manually or Excel aided we get:
For a+d=8 and c+d=1 the expression is multiple of 41 (41*4)
And the same is true for the pairs (a+d, c+b)=(7,6) (6,11) (5,16) (13,17) (14,12) (15,7) (16,2)
So this determine the form of the numbers.
Verifying: For ex: 7986
7986+6897=14883 (41*363)
To the puzzle: 2016 is a+d=8 c+d=1. The next number with this form will be 2106.
But there is a closer number to 2016, with the form a+b=7 c+d=6. The number is 2065.
Edited on June 26, 2016, 3:17 pm
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Posted by armando
on 2016-06-26 14:26:05 |
Better
