Everyone loves going for vacation. It is always a joy to visit new places, enjoying how they are different from where we live. They may be a hill station, or a desert or even a remote village. Once there was a man called Gulliver lived in 17^th century (remember we are at 21^st century) who love traveling to different places. He was a doctor, but he also wished to explore new places.

To fulfil his ambition he joined as a doctor in a ship, on those days ships were the only mode of transport for log travel, no flight or bus. In addition, they did not had a proper map or globe to identify all the countries in the globe. They used to travel by ship to explore new countries.

Once on their journey the ship broke down and Gulliver landed on island called Lilliput. The people in the island were tiny in size; as such, a group of people can be place on Gulliver’s hand. They were initially scared of him and soon he earned the trust of the people, they started treating him as guest. After few days, he felt bored and wanted to move back to London. The Lilliput king was not ready to release him. Read out this interesting book Gulliver’s Travels by Jonathan Swift. There are different versions available for to cater young readers

Hello this time I will discus about Quick divisibility tests which will be useful in finding the factors of a
number. This is basically required to solve questions involving L.C.M and G.C.F of numbers .
Checking of whether a number is exactly divisible by 7 , 13 ,19 ,23 , 27,29,31 is difficult by using
multiplication table . Quick divisibility tests sequentially starting from 2 are discussed below which
saves a lot of time while finding factors .
1. Test for 2 – If unit digit is 0,2,4,6,8 2 the number is divisible by 2 . ( even numbers )
2. TEST for 3 – If the sum of digits of a number is divisible by 3 .
EX- 12345 ,
sum of digits is 1+2+3+4+5 = 15 , which is divisible by 5 .
So the number 12345 is divisible by 3 .
3 . Test for 4 – If the number formed by last 2 digits is divisible by 4 .
EX- 123456 ,
the number formed by the last 2 digits is 56 , which is divisible by 4 .
So 123456 is divisible by 4
4 . Test for 5 – Unit digit is 5 or 0 .
5. Test for 6 – If it is divisible by both 2&3 .( Number is even and sum of digits is divisible by 3)
6. Test for 7 –
(1) 3 digit number –
Multiply the unit digit by 2 and subtract it from the rest . If the result is divisible by 7 ,whole number
is divisible by 7 .
Ex- 945 , unit digit is 5 and rest is 94 ,
94 – 2×5 = 84 , which is divisible by7 ,
So 945 is divisible by 7 .
(2) For numbers having minimum of 4 digits –
If q is number formed by last 3 digits and p is number made up of rest of the numbers ,
then if ( p-q ) is divisible by 7 , then the whole number is divisible by 7 .
Four digit number – 9331 ,q= number formed by last 3 digits = 331 , p = rest of the numbers = 9
p – q = 9 – 331 = – 322 ,which is divisible by 7
Hence 9331 is divisible by 7 .
Five digit number – 34426 , p=34 , q =426 ,
p – q = 426 – 34 = 392 which is divisible by 7
( since it is now a 3 digit number you can test it as discussed above , 39 – 2×2=35, divisible by 7 )
Hence 34426 is divisible by 7 .
Six digit number – 976213 , p = 976 , q = 213 ,
p – q = 976 -213 = 763 which is divisible by 7 .
So the number 976231 is divisible by 7 .
7 . Test for 8 – If the number formed by last 3 digits is divisible by 8 .
Ex- In 123456
the number formed by last 3 digits is 456, which is divisible by 8.
So 123456 is divisible by 8
8 . Test for 9 – If sum of digits is divisible by 9 .
9 . Test for 10 – If unit digit is 0
10 . Test for 11 –
If the difference of the sum of the digits in the even places and the sum of digits in odd places is
divisible by 11 or 0 .

Ex – 58432 , numbers in odd places are 5,4,2 and numbers in even places are 8,3
( 5+4+2 ) – ( 8+3 ) = 11 – 11 = 0 ,
so 58432 is divisible by 11 .
11 . Test for 12 –If it is divisible by 4&3 . Number formed by last 2 digits is divisible by 4 and sum of
digits is divisible by 3 .
EX- In 123456 ,
sum of digits is 1+2+3+4+5+6 = 21 , divisible by3 ,
number formed by last two digits is 56 , divisible by 4 .
So 123456 is divisible by 12 .
12 . Test for 13 —
Method 1 – Multiply unit digit by 9 and subtract it from rest . If the result is divisible by 13 the whole
number is divisible by 13 .
Ex- IN 559 , unit digit is 9 and rest is 55
55 – 9×9 = 55 – 81 = – 26 ,which is divisible by 13 .
So 559 is divisible by 13
Method 2 — Multiple unit digit by 4 and add it to the remaining number .
Ex – 12519 , unit digit is 9 and rest is 1251
1251+ 4×9 = 1251 + 36 = 1287 ,
Repeat the same method for 1287
128 + 4×7 = 128+28 =156 , which is divisible by 13 .
So 12519 is divisible by 13
13 . Test for 19 – Multiply unit digit by 2 Add to the remaining .
Ex – 65341 , unit digit is 1 and rest is 6534
6534 +1×2= 6536 , ( Repeat the same process again and again )
653 + 2×6 = 665 ,
66 + 2×5 = 76 , divisible by 19 .
So 65341 is divisible by 19
14 – Test for 23 – Multiply unit digit by 7 and add to the remaining .
Ex – 15019 , unit digit is 9 and rest is 1501
1501+ 9×7 = 1501+63 =1564 ,
Again 156+4×7 = 156+28 = 184 ,
18+4×7 =46 is divisible by 23 .
So 15019 is divisible by 23 .
15 . Test for 27 –
(1) For 3 digit number — Multiply the 1 st digit by 8 , subtract it from rest .
Ex – 675 , 75 – 6×8 = 75 – 48 =27 which is divisible by 27 .
So 675 is divisible by 27
( 2 ) Digits more than 3 – If last 3 digits is “b” and rest is “ a “ , a+b is divisible by 27
then the whole number is divisible by 27 .
Ex– In 12312 ,
b =312, a=12 ,
a+b = 312+12=324 which is divisible by 27
So the number 12312 is divisible by 27
16 . Test for 29 – Multiply unit digit by 3 and add to the remaining .
Ex – 10324 , unit digit is 4 and remaining is 1032

1032 + 3×4 =1044 , ( repeat the same again and again )
104+3×4 =116 ,
11+3×6=29 , which is divisible by 29 .
So 10324 is divisible by 29
MAGICAL NUMBER 37 ———– Here I will discuss some curious
properties of 37 along with its divisibility test .
A 3 digit number with all digits same when divided by the sum of digits , the result is always 37 .
111 / ( 1+1+1) = 111 / 3 = 37 or 37 x 3 = 111
222 / (2+2+2) = 222 / 6 = 37 or 37 x 6 = 222
…………………………….. ——————
999 / ( 9+9+9) = 999 / 27 = 37 or 37 x 27 = 999
CURIOUS PROPERTIES OF 37
(1) The sum of it’s digits x 37 = The sum of the cubes of it’s digits .
( 3 + 7 ) x 37 = 3 3 + 7 3
(2) The sum of the squares of it’s digits minus the product of it’s digits is 37 .
( (3 2 + 7 2 ) – (3 x 7) ) = 37
(3) Take a 3 digit multiple of 37 ,
for example 37 x7 = 259 ,
then an end-around carry produces 925 , 592 both divisible by 37
DIVISIBILITY BY 37 – Separate the number from right to left into groups of 3 digits . ( the
last may be incomplete ) . Consider each group an independent number. Add the numbers . If the
number is divisible by37 ,the entire number is divisible by 37 .
EXAMPLE – 5254 ,
two groups are 5 ,254 .
254+5 = 259 , 259 / 37 = 7 ,divisible by 37 ,
Hence the 5254 is divisible by 37

Everyone knows a chameleon can change color but do you know it has a tongue as long as its body. Unlike us where our both the eyes move together in same direction, Chameleon can move its eyes in different direction. All these funny facts about animals present in the book named “Ten must read animal stories” by Ruskin Bond.

There is a story about a boy Kishore who is curious to see a tiger and waits awake for a whole night. Another story is about a parrot who refuses to speak even after several rounds of speech therapy and the owner starts yelling at it. The story of Hornbill (a bird if you don’t know google it, a very beautiful one) which is saved from predator. Another interesting story about Hare, Jackal, Otter and a monkey is there.

There are also many other stories with different animals as main character. The stories are very simple and easy to read. I’m sure reading it for children’s will delight them. Happy reading.

Copyright is a legal instrument that protects original works of authorship. These works include any type of artistic, literary, musical, or dramatic creation such as books, poetry, movies, music, lyrics, computer software, and architecture. After writing music or lyrics, a composer or songwriter can register a copyright that will protect this intellectual property. The copyright remains in effect for 70 years after the death of the artist. The Copyright Act of 1976 was instrumental in outlining a number of rules that pertain directly to music law and copyright. The person who writes a song owns the legal right. A song writer can sell the rights or they can allow someone else to perform the song while they get some of the profits. The creator of musical work gets a copyright for the work; people who want to broadcast a work or perform it live usually have a license from the owner. Everyone who is interested in the music industry is subject to laws that govern their rights and actions. Music publishers must abide by these laws, as must artists, producers, promoters, and executives. These laws pertain to the terms of record deals, print licensing for sheet music, royalties for artists’ public performances, and more. The Copyright allows artists to reap the economic benefits of their creative endeavours by enabling creators, producers, publishers and distributors of artistic work to control how, when and whether their works are used. For music, the key concerns are determining who owns the copyright in newly created songs and understanding what right and benefits are given to copyright owners. Therefore, when your children’s compose music or write lyrics on their own, make sure they are copyrighted and they own their legal rights. With Regards Mrs. Arokia Anitha (Advocate & IP Legal Advisor) Parent of HANIRVAN – LKG – D

In today’s technological world, everything have turned to be digital. A day is not spent
without smart phones, laptops, ipads, credit / debit cards, online shopping, etc,. In
India, all cyber related issues are governed by Information Technology Act, 2000.
Cyber law is the area of law that deals with the Internet’s relationship to technological
and electronic elements. The digital education and use of internet have started from
kindergarten and there is no age limit. Hence, everyone should be aware that, usage of
internet provides knowledge and benefit and at the same time it can also put you in
trouble if not accessed in the right way.
We should be very careful while accessing internet or sending messages and mails in
social media like facebook, twitter, whatsapp etc. Even a small mistake can put you in
big legal trouble if it’s not used in proper manner.
Certain precautionary measures should be taken, especially by parents while your
children are using the internet. It is always the parents who have to keep a watch on the
sites that your children are accessing, to prevent any kind of harassment or depravation
in children.
We should guide our children to avoid disclosing any personal information to strangers
via e-mail or while chatting. One must avoid sending any photograph to strangers by
online as misusing of photograph incidents increasing day by day. Parents should avoid
sharing debit /credit cards to their children and also educate them not to share the
details to any strangers while using internet.
Parents should make sure that the internet should majorly be used by your kids to
develop their knowledge, skill, communication and not to put them in any kind of
trouble.

   
1. Test for 19 – Multiply unit digit by 2 Add to the remaining. Ex – 65341 , unit digit is 1 and rest is 6534 6534 +1×2= 6536 , ( Repeat the same process again and again ) 653 + 2×6 = 665 , 66 + 2×5 = 76 , divisible by 19 . So, 65341 is divisible by 19
2. Test for 23 – Multiply unit digit by 7 and add to the remaining. Ex – 15019 , unit digit is 9 and rest is 1501 1501+ 9×7 = 1501+63 =1564 , Again 156+4×7 = 156+28 = 184 , 18+4×7 =46 is divisible by 23 . So, 15019 is divisible by 23.
3. Test for 27 – (1) For 3 digit number — Multiply the 1st digit by 8 , subtract it from rest . Ex – 675 , 75 – 6×8 = 75 – 48 =27 which is divisible by 27 . So 675 is divisible by 27 ( 2 ) Digits more than 3 – If last 3 digits is “b” and rest is “ a “ , a+b is divisible by 27 then the whole number is divisible by 27 . Ex– In 12312 , b =312, a=12 , a+b = 312+12=324 which is divisible by 27 So the number 12312 is divisible by 27
4. Test for 29 – Multiply unit digit by 3 and add to the remaining. Ex – 10324 , unit digit is 4 and remaining is 1032 1032 + 3×4 =1044 , ( repeat the same again and again ) 104+3×4 =116 , 11+3×6=29 , which is divisible by 29 . So, 10324 is divisible by 29
Kiran Bala Dash (Grandmother of Tanay – 4E and Vinay – LKG D).
Retired Lecturer in Mathematics

Danny was a smart young boy. When he was 4 months old he has lost his mother. He was brought by his dad with utmost care. They owned a Gas station and a mechanic workshop. They didn’t have a house and stayed in a caravan. The caravan doesn’t have electricity and they always light candles in the evening. His dad was a fun-loving person and treat Danny as a friend. Danny at a young age instead of playing with toys was spending time at his dad’s workshop and learned about repairing the car.

All went fine until a day when Danny came to know about the secret habit of dad. Because of which his dad has got into a trouble and story starts on a high pitch with brave adventures of Danny on saving his dad. Usually, a dad would help his son to achieve his Goals. Things are different here as Danny the world champion helps his dad to achieve his passion. As always the author Roald Dahl entertains us with the beautiful story of a child’s brave adventure and a lovely relationship between Father and Son. Enjoy the book “Danny the Champion of the world” by Roald Dahl. Happy reading.

Divisibility Tests – Continued (Part II)

1. Test for 8 – If the number formed by last 3 digits is divisible by 8 . Ex- In 123456                                                                                                  the number formed by last 3 digits is 456, which is divisible by 8.  So, 123456 is divisible by 8
2. Test for 9 – If sum of digits is divisible by 9.
3. Test for 10 – If unit digit is 0
4. Test for 11 –  If the difference of the sum of the digits in the even places and the sum of digits in odd places is divisible by 11 or 0 .           Ex – 58432 ,   numbers in odd places are    5,4,2    and  numbers  in even places are 8,3( 5+4+2 ) – ( 8+3 ) = 11 – 11 = 0 ,                              so 58432 is divisible by 11 .
5. Test for 12 –If it is divisible by 4&3. Number formed by last 2 digits is divisible by 4 and sum of digits is divisible by 3.                        EX- In 123456  ,  sum of digits  is 1+2+3+4+5+6 = 21 , divisible by3,  number formed by last two digits is 56 , divisible by 4 .                        So, 123456 is divisible by 12.
6. Test for 13 — Method 1 – Multiply unit digit by 9 and subtract it from rest. If the result is divisible by 13 the whole number is divisible by 13.  Ex- IN  559, unit digit is 9 and rest is 55     55 – 9×9 = 55 – 81 = – 26, which is divisible by 13.  So 559 is divisible by 13                                                  Method 2 — Multiple unit digit by 4 and add it to the remaining number.    Ex – 12519 , unit digit is 9 and rest is 1251                                     1251+ 4×9 = 1251 + 36 = 1287 ,  Repeat the same method  for 1287                                                                                                                               128 + 4×7 = 128+28 =156 , which is divisible by 13  .  So, 12519 is divisible by 13

Kiran Bala Dash (Grandmother of Tanay – 4E and Vinay – LKG D).

 Retired Lecturer in Mathematics