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