Remainders -4 Procedure for finding the numbers to get the remainders

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.

Remainders-3

You can have problems with 4 conditions also-
Example-Remainder 3 on division by 5
                  -do-      4      -do-          9
                  -do-      11     -do-         13
                  -do-      12      -do-        17
Try doing this as per method given by me.
Do you find anything in the problem which will enable you to simplify the method?         

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

Arithmetical progressions-2

In my previous blog,I had introduced the subject.
I referred to a progression with the first term 'a' =32 and common difference 'd'=5.
Here is how one can find subsequent terms-
For the second term you have to add the value 'd=5' once to 'a=32' giving you 32+5=37.
For the third term you have to add twice the same value giving you 32+10 =42.
For the fourth term you have to add thrice giving you 32+15=47 and so on.
Hence for the 16th term you have to add the value 15 times to give you 32+15*5=32+75=107
For the 45th term you have to similarly add the value 44 times to give you 32+44*5=32+220=252.
Now for finding the sums-
If the total no of terms is odd ,there will always be a middle term ,double the value of which will equal the total  of the value of the first term and the value of the last term.The total of the value of the second term and the penultimate term will be the same amount..This process continues throughout.
When there are 45 terms in all ,the middle term will be the 23 rd term viz 32+22*5=32+110=142.Double this value is 284.
The total value of the first  and 45th terms is  (32+252)=284..Similarly the total of the values of  the second term (37) and 44th term (247) is (37+247)=284..Similarly you get the same result considering the third term(42) and 43rd term (242) viz (42+242)=284 and so on..
The total of all the 45 terms will hence be 45 times the value of the middle term viz 45*142=6380.
You can check the correctness of this method by first  considering 3 terms then 5 terms,then 7 terms and so on
When the total no of terms is even there will be two middle terms and hence you have to take the mean value of those two terms.
When there are 16 terms,the 8th term(67) and the 9th term(72) will be the middle terms and hence you have to take their mean and multiply by 16,to get the total of all the 16 terms.You will have 16*(67+72)/2=8*139=1112 .

Arithmetical Progressions

I wish to familiarise the topic Arithmetical Progressions
Any sequence of numbers starting with a particular number and progressively increasing with equal additions can be stated to be forming such progressions.
The start number can be called the first term-generally denoted by 'a'
The additions in equal values can be called the 'common difference-generally denoted by 'd'
'a' can be anything like 5,    100      ,3/7    ,4.05,    6^2   etc
'd' can also be similarly anything-can also have negative values.
A few examples shown below-
100,103,106,109,112......
(4.5)   .(5.2)    (5.9)    (6.6)....
7/45......9/45    11/45...13/45....15/45    17/45
1000,992,984,976,968...  etc
Given the first term  like 32 and common difference like 5,we have methods to find out the 16th or 45th etc terms.Also the sums of the various terms upto the 16th term or 45th term.
I will explain the methods in my next blog.

Remainders

I am posing a question for you to solve.
I want you to find the smallest number which gives a remainder 2 when divided by 3,a remainder 3 when divided by 5 and a remainder 5 when divided by 7., 
How can you find more such numbers?