 Welcome friends. So in this QoD, so this was posted as QoD Now in this question of the day problem, it's given that what are the last two digits two digits in the number 11 to the power 1 1 1 So this is what we have to find out. So last two digits for example, if 574 the last two digits is seven and four So if we take this example, then let us say if I have to find out 11 square It is 121. So last two digits is 2 1 11 to the power 3 is 1 3 3 1 so last two digits are 3 and 1 So likewise, we have to find out what is 11 to the power 1 1 1 last two digits Okay, last two digits now Usually the tools which are used for such problems are either binomial theorem so binomial theorem or something called modular arithmetic modular Arithmetic so but in this case You'd assume that you guys don't know either of these tools. So how to solve such problems. So now Let us say it. So what I have expressed here as if you can see 10 plus 1 to the power n so 11 can be always expressed as 11 is nothing but 10 plus 1 so if I raise any power to 11, which is equal to raising 10 plus 1 whole to the power n Okay, this is what the basics of binomial theorem also suggests But we will not go into the deeper details of it We'll try to solve this problem without going into the details of binomial theorem for that We need to do some trend analysis Let us say 11 to the power 2 was how much Can be written as 10 plus 1 squared so hence if you see it is nothing but by our a plus b whole square identity We'll get 10 square plus 2 times 10 now. I'm putting a dot here. Don't confuse this with decimal This is not decimal. This is a product sign So 2 times 10 plus 1 squared correct next 11 to the power 3 and 11 to the power 3 is nothing but 10 plus 1 to the power 3 and you know the identity a plus b whole cube and You will get 10 cube plus 3 times 10 square Plus 3 times 10 plus 1 So I'm not writing 1 cube here. I'm just writing 1 now 11 to the power 4 Whether you know the identity or not I'll tell you if by just by multiplying these two twice you'll get this 10 to the power 4 plus 4 times 10 cube Plus 4 sorry 6 times 10 squared Plus 4 times 10 plus 1 so if you notice the trend as you are You know the power as you are increasing the power 2 3 and 4 so you see a trend here. That is 2 This is 2. This is 3 This is 4 likewise here if you see This is 3 This is 4 right the last Second last term has The second and the second last terms has this these coefficients So likewise if you if you just try to extrapolate so 11 to the power 5 will be 10 plus 1 to the power 5 is equal to 10 to the power 5 plus, you know the next term 5 into 10 to the power 4 Right plus whatever it is these are our powers of 10 the second last term definitely will be 5 times 10 plus 1 Right, so hence if you keep on doing this exercise so for 11 to the power 1 1 1 this will be equal to 10 plus 1 to the power 1 1 1 so hence it is equal to 10 to the power 1 1 1 Plus 1 1 1 times 10 to the power 1 1 0 1 less than if you see If it is 4 this one is 3 if it is 3 this one is 2 if it is 5 this one is 4, isn't it? So if it is 1 1 1 here, it will be 10 to power 1 1 10 though It is not important for this problem, but for understanding purposes So now last second last term will be 1 1 1 till times 10 plus 1 and mind you before this term There will be power of 10 Yeah, so all here third last is power 10 if you power 2 can you see this? Power 2 So it will power 2 here also will be power 2 so hence if you add these terms The last two digits is of obviously 0 and 0 why because all these are higher powers of 10 Now that means the total 11 to the power 11 1 1 the last two digits will come from this sum and this sum is Nothing, but 1 1 1 and 1 Okay, so even if after adding let us say Whatever whatever the sum comes here you add this to this so you have some Digits here and last two digits are 0 0 for this part and for this part you have 1 1 1 1 so if you add both of them you'll get 1 1 and whatever digits are there it is immaterial right so what do we get the last two digits are nothing but 11