Finding last two digit of a number

sorry folks! it has been a long time since I posted fundas (busy in my new coaching job). anyways here is a funda to cover up my long pending quota

I am dividing this method into four parts and we will discuss each part one by one:

a. Last two digits of numbers which end in one
b. Last two digits of numbers which end in 3, 7 and 9
c. Last two digits of numbers which end in 2
d. Last two digits of numbers which end in 4, 6 and 8

Before we start, let me mention Binomial theorem in brief as we need it for calculation

(x+a)^n = Nc0*a^n+Nc1*a^n-1*x+Nc2*a^n-2*x^2+….. where NcR =n!/(r!(n-r)!)

a. Last two digits of numbers which end in one

This is best explained with an example.

What are the last two digits of 31^786?

Solution: 31^786 = (30+1)^786 = 786C0*1^786+786C1*1^785*30+786C2*1^784*30^2…..

Note that in the above expression, after the first two terms, all the terms will have two or more 0’s at the right hand side (since the expression is multiplied by 30^x). Hence to find the last two digits, only first two terms needs to be evaluated. First term equals to 1 and second term equals to 786*30 = 23580 hence last two digits will be 81 (which is last two digits of 23580 and +1 hence 80+1)

Wiyee!! That’s really long isn’t it? Well those who need a shortcut of this read on!!

Multiply the tenth digit ( 3 in the above example) with the unit digit of the exponent (6 in the above example) to get the tenth digit. Unit digit is always one when a number ends in 1.

Now it’s a two second job isn’t it??

Few exercises for you:

41^2789

71^56747

51^456*61^567

Post your answers using comment option

Now I have covered 25% of this exercise i.e. part A alone. Based on your response, if you find this useful, I will go ahead and post the approach for below kind of numbers

b. Last two digits of numbers which end in 3, 7 and 9
c. Last two digits of numbers which end in 2
d. Last two digits of numbers which end in 4, 6 and 8

Thanks,

Quant-Master