83516560061 என்ற எண் 19 ஆல் மீதமின்றி வகுபடுமா? (www.vedic-maths.in)

[Anbazhagan Devaraj]

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83516560061 என்ற எண் 19 ஆல் மீதமின்றி வகுபடுமா? (www.vedic-maths.in)

எண்ணின் கடைசி இலக்கத்தை இருமடங்காக்கி (Double the last digit) கடைசி இலக்கம் தவிர்த்த எண்ணுடன் கூட்டவும் (சிறிய எண்ணாக வரும் வரை, இதையே பல முறை செய்யவும்). கடைசியாக வரும் விடை 19 எனில், அந்த எண் 19 ஆல் மீதமின்றி வகுபடும்.


உதாரணம் 1: 76 divisible by 19?

76-> 7 + (6X2) -> 19, எனவே 76 ஆனது 19 ஆல் மீதமின்றி வகுபடும்.


உதாரணம் 2: 228 divisible by 19?

228-> 22 + (8X2) -> 38; 3 + (8X2) -> 19, எனவே 76 ஆனது 19 ஆல் மீதமின்றி வகுபடும்.


உதாரணம் 3: 551 divisible by 19?

551-> 55 x (1X2) -> 57; 5 + (7X2) -> 19, எனவே 551 ஆனது 19 ஆல் மீதமின்றி வகுபடும்.


உதாரணம் 4: 1102 divisible by 19?

1102-> 110 x (2X2) - > 114; 11 x (4x2) -> 19, எனவே 1102 ஆனது 19 ஆல் மீதமின்றி வகுபடும்.


உதாரணம் 5: 83516560061 divisible by 19?

83516560061->

8351656006 x (1X2) - > 8351656008;
835165600 x (8X2) - > 835165616;
83516561 x (6X2) - > 83516573;
8351657 x (3X2) - > 8351663;
835166 x (3X2) - > 835172;
83517 x (2X2) - > 83521;
8352 x (1X2) - > 8354;
835 x (4X2) - > 843;
84 x (3X2) - > 90;
90 ஆனது 19 ஆல் வகுபடாது, எனவே 83576560061 ஆனது 19 ஆல் மீதமின்றி வகுபடாது.

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[mskmoorthy]

Let me give a short justification why the above method works in terms of mathematics that one may be familiar with in high school.

We want to show that 10x+y is divisible by 19 iff x+2y is divisible by 19.

10x+y is divisible by 19 iff (-9)x+y is divisible by 19 (10 divided by 19 gives a reminder of 10 or equivalently -9)

(Adding) Hence 10x+y +-9x+y is divisible by 19 iff 10x+y is divisible by 19
i.e., x+2y is divisible by 19 iff 10x+y is divisible by 19.

You can also calculate the remainder using this method. If the remainder during the final division is even number 2r then the remainder is r. If it is odd s then the remainder is (19+s)/2
(You can understand why we divide by 2 by the above argument). For example, in the last case of post number 1
83516560061 divisible by 19 -it is not divisible by 19 remainder will be
calculate the remainder of the last step 90/19 which will be 14 - an even number and hence the remainder is 7.

 

[mskmoorthy]

In my previous post there is a mistake in calculating the remainder for the problem posted in post no 1 - The correct remainder is 5 - It needs a bot more calculation.

My posts are meant to relate to what the students learn in high school. The suggestions may help them to explore on their own and develop/extend the ideas. My comments are not meant for anything else (kindly do not misunderstand my intentions).

Now we can extend this method to divide by 29,39,49, 59 etc.

10x+y is divisible 29 , then we can add 3 times to get 30x+3y which will also be divisible by 29.
30x+3y is disible by 29 implies x+3y is divisible by 29.
if x+3y is divisible by 29, then 30x+3y is divisible by 29, which 10x+y is divisible by 29.

for 39 it will be x+4y, for 49 it will be x+5y and so on

This may encourage children to explore further in mathematics and find fun and happiness!!

 

[Anbazhagan Devaraj]

Test for Divisibility by 5
- Check if the last digit is 0 or 5
Example: 123512341325
Since the light digit is 5, the number is divisible by 5

Test for Divisibility by 2
- Check if the last digit is even (0, 2, 4, 6, 8)
Example: 12312348
Since the last digit is 8, the number is divisible by 2

Test for Divisibility by 4
- Check if the last 2 digits is divisible by 4
Example: 123124124
Since 4|24, the number is divisible by 4

Test for Divisibility by 8
- Check if the last 3 digits is divisible by 8
Example: 9798743648
Since 8|648, the number is divisible by 8

Test for Divisibility by 16
- Check if the last 4 digits is divisible by 16
Example: 7849758432
Since 16|8432, the number is divisible by 16

Test for Divisibility by 11
- By subtracting the sum of all the even position of the digit by the sum of all the odd position of the digit
Example: 19203182
Odd sum: (1 + 2 + 3 + 6) = 12
Even sum: (9 + 0 + 1 + 2) = 12
Difference is 0 and 0 is divisible by 11. Hence the digit is divisible by 11...