Regarding the problem mentioned in my blog Remainder-3 the solution is found as follows-
For remainder 12 in respect of division by 17,the numbers are 29,46,63...
The number 63 satisfies the condition-remainder 11 for division by 13.
For remainder 4 for division by 9,we add 17*13=221 continuously to 63 and test.
The numbers are 63,284,505,726,947,1168,1389,1610,1831..
The number 1831 satisfies remainder 4 for division by 9.
To consider division by 5(Remainder3),we add 17*13*9=1989 continuously to 1831.
The numbers are 1831,3820,5809,7798...
The number 7798 satisfies a remainder 3 for division by 5.
Hence 7798 is the number which satisfies all 4 conditions as required.
The method adopted can be simplified taking into account the fact below-
Instead of the remainder in each case,we consider how much the numbers are short for an even division by the dividing number.
For 13,the remainder should be 11 and for 5 the remainder should be 3-in both case the number should be 2 short for an even division by 13 and 5 and so the smallest such number is 13*5-2=63.
For 17 and 9 ,similarly the number should be short by 5(to give remainders 12 and 4) and so the smallest such number is 17*9-5=148
This number 148 also gives remainder 3 for division by 5.
We now only have to satisfy the condition of remainder 11 for division by 13.
We hence add 17*9*5=765 continuously to 148 and test.
The numbers we get by adding 765 are 913,1678,2443,3208,3973,4738,5503,6268,7033 and finally7798 which is what we found earlier.
For remainder 12 in respect of division by 17,the numbers are 29,46,63...
The number 63 satisfies the condition-remainder 11 for division by 13.
For remainder 4 for division by 9,we add 17*13=221 continuously to 63 and test.
The numbers are 63,284,505,726,947,1168,1389,1610,1831..
The number 1831 satisfies remainder 4 for division by 9.
To consider division by 5(Remainder3),we add 17*13*9=1989 continuously to 1831.
The numbers are 1831,3820,5809,7798...
The number 7798 satisfies a remainder 3 for division by 5.
Hence 7798 is the number which satisfies all 4 conditions as required.
The method adopted can be simplified taking into account the fact below-
Instead of the remainder in each case,we consider how much the numbers are short for an even division by the dividing number.
For 13,the remainder should be 11 and for 5 the remainder should be 3-in both case the number should be 2 short for an even division by 13 and 5 and so the smallest such number is 13*5-2=63.
For 17 and 9 ,similarly the number should be short by 5(to give remainders 12 and 4) and so the smallest such number is 17*9-5=148
This number 148 also gives remainder 3 for division by 5.
We now only have to satisfy the condition of remainder 11 for division by 13.
We hence add 17*9*5=765 continuously to 148 and test.
The numbers we get by adding 765 are 913,1678,2443,3208,3973,4738,5503,6268,7033 and finally7798 which is what we found earlier.
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