Remainders-2

There has been no response to my previous blog.I desired to know the smallest number which gives remainder 2 when divided by 3,remainder 3 when divided by 5 and  remainder 5 when divided by 7.
There is a method known as chinese remainder theorem.,But I will give you another method
First take the largest divider(7) and the remainder(5).The first number which satisfies the requirement is 7+5=12.,Further numbers can be found by adding 7.They are thus 12,19,26,33,40,47,54,61,68,75,82.....etc
Check these whether you get remainder 3 when divided by 5.Only 68 will satisfy this requirement.
This number 68 surprisingly satisfies the third condition of giving reminder 2 when divided by 3.
So 68 is the required number.
Suppose I had altered condition that you must have a remainder 1 when divided by 3.You then have to add 7*5=35 to 68 repeatedly to check whether you get remainder 1 when divided by 3.The first number you get by adding 35 to 68 is 103.This satisfies the condition that you get remainder 1 when divided by 3
So 103 is the required number which satisfies all the 3 conditions.. .
To get further numbers you have to add 7*5*3=105 repeatedly to 103 .You will have the numbers 208,313,418,523 .....etc.
All these will satisfy all the 3 conditions.
To see whether you have understood the method,I will give you a more difficult problem.
You must find the smallest number which gives remainders as follows-
Remainder 5 when divided by 13
....do-       8     ...do-.             17
...do.....    11.....do-................23
You must get the answer 2150.
Further numbers obtained by adding 13*17*23=5083 repeatedly to 2150.
They are 7233,12316... etc