# Divisibility Problems Tips and Tricks

## Divisibility Tips and Tricks to solve Questions Quicker

Always to check if x divides a number break it into

• Number, N = q(x) + r
• where q is quotient and r remainder and x can be said as divisor

There are three numbers a, b and c. You have to find out the largest number x that divides all the 3 numbers.

Way to solve Type 1 Question

Now a is the greatest number, b is the 2nd largest of the three and c is lowest.

Find the HCF of (a – b), (b – c) and (a – c)

e.g. The greatest number that will divide 63, 138 and 228 so as to leave the same remainder in each case:

a = 228, b = 138 and c = 63

HCF of ((228-138),(138-63),(228-63))
HCF of (90,75,165)
75=3*5*5
90=2*3*3*5
165=3*5*11

Thus HCF is 3*5 = 15

When there is a number is N, when divided by another number a, the remainder is r1. When the same number is divided by a number b, then remainder is r2. What is the number ?

Clearly,

• N = x(a) +r1, where x is natural number
• N = y(b) + r2,where y is a natural number.

Lets see with an example –

What is the smallest four number which when divided by 6, leaves a remainder of 5 and when divided by 5 leaves a remainder of 3?

• N = 6x +5 and N = 5y + 3
• so 6x +5 = 5y + 3
• => y-x = (x+2)/5

Thus a+2 should be divisible by 5 smallest value would be x =3.

• so y – 3 = (3+2)/5 => y = 4
• Thus N = 6x +5 = 23

Now to calculate largest 4 digit number –

• Find remainder by dividing 1023/23 => remainder 11 as (smallest 4 digit number + N)/ N = Remainder 11.
• Number = 1023 – 11 = 1012.

Now in the same Question if you had to find largest 4 digit number then same thing but new equation

The number less than 1000 exactly divisible by 6 is 996. And 995 is the number exactly divisible by 5 and less than 1000.

• Let the number be x
• x = 996+6p+5 and x = 995+5q+3
• => 996+6p+5 = 995+5q+3
• => 1001+6p = 998+5q
• => 3+6p = 5q
• the lowest integer pair satisfying this equation would be (p,q) = (2,3)
• => x = 996+6*2+5 = 995+5*3+3 = 1013.

Greatest power of p that will divide N!

This can be solved directly by a formula –

[N/p] + [N/p2] + [N/p3] …..

e.g. for 40! and 5

[40/5] + [40/52] + [40/53] …..

8 + 1 + 0 = 9