Number System

[prafull26]

there is a question on compound interest.

a certain amount has Rs 1000 as interest after 1 year.
after 3 years the compound interest is Rs 3310. what is the rate of interest.

there should be a shortcut to find out the rate of interest.

answer is 10%. other options are 15 , 25 , 20.

[utsavmamoria]

Originally Posted by prafull26

there is a question on compound interest.

a certain amount has Rs 1000 as interest after 1 year.
after 3 years the compound interest is Rs 3310. what is the rate of interest.

there should be a shortcut to find out the rate of interest.

answer is 10%. other options are 15 , 25 , 20.

If you just want a short method , use a litle bit of observation

note the interest 3310
1331 is the cube of 11 or (1.1)^ 3 = 1.331

1.331 x 10000 = 13310

so in th eproblem we can say that the intial principal is 10000
rate is 10% and interest in 3 years is 3310

But I am not sure that there should be a shorter more methodical approach to this

[reachmonil]

Q. What are the last 2 digits in the expansion of 2^999.

I solved this one using Ferment's Little theorem and Euler's Inverse Theorem.
This question then simplifies to 2^-1 mod 100.
Using Inverse Theorem,
2x mod 100 = 1
(100k + 1) / 2. But, getting stuck here. Can anyone shed some light?

[krsh.vik]

Originally Posted by reachmonil

Q. What are the last 2 digits in the expansion of 2^999.

I solved this one using Ferment's Little theorem and Euler's Inverse Theorem.
This question then simplifies to 2^-1 mod 100.
Using Inverse Theorem,
2x mod 100 = 1
(100k + 1) / 2. But, getting stuck here. Can anyone shed some light?

we should find 2^999 mod 100
i.e., 2^997 mod 25
now 5 and 2 are coprimes

E(25)=20
2^980 mod 25 = 1
2^17 mod 25 = 7^3*4 mod 25
= 49*3 mod 25
= 22

answer 88

[prade]

Originally Posted by reachmonil

Q. What are the last 2 digits in the expansion of 2^999.

I solved this one using Ferment's Little theorem and Euler's Inverse Theorem.
This question then simplifies to 2^-1 mod 100.
Using Inverse Theorem,
2x mod 100 = 1
(100k + 1) / 2. But, getting stuck here. Can anyone shed some light?

2^999 mod 100 = 2^997 mod 25 multiply by 4 later
EN(25) = 20

2^17 mod 25 = 2^-3 mod 25 multiply by 4 later
8x = 1 mod 25
=> 8x = 25y +1
we can say x = (25y+1)/8 => (y+1) mod 8
hence y = 7 which will ensure it si divisible by 8
=> x = 22

Multiply by 4 to get the complete answer of 88

[reachmonil]

Originally Posted by prade

2^999 mod 100 = 2^997 mod 25 multiply by 4 later
EN(25) = 20

2^17 mod 25 = 2^-3 mod 25 multiply by 4 later
8x = 1 mod 25
=> 8x = 25y +1
we can say x = (25y+1)/8 => (y+1) mod 8
hence y = 7 which will ensure it si divisible by 8
=> x = 22

Multiply by 4 to get the complete answer of 88

But, I am a little amazed at why we can't get the answer using the same Inverse formula without dividing by 4. Does that mean, the 2 numbers need to be co-prime? Nay, I don't thin so.

[hameed]

Originally Posted by reachmonil

But, I am a little amazed at why we can't get the answer using the same Inverse formula without dividing by 4. Does that mean, the 2 numbers need to be co-prime? Nay, I don't thin so.

I find the easiest way for any such prblm to be ...

2^999 = (2^10)99*2^9
last 2 digits of (2^10)99 is 24 [ 2^10 raised to even power ends with 76, and odd power ends with24]
and last 2 digits of 2^9 is 512

So last 2 digits is 12*24 = x88

Rock on!
hameed

[prade]

Originally Posted by reachmonil

But, I am a little amazed at why we can't get the answer using the same Inverse formula without dividing by 4. Does that mean, the 2 numbers need to be co-prime? Nay, I don't thin so.

Monil yeh step kaise aaya...
2^999 mod 100 = 2^-1 mod 100

Note:2 and 100 are not co prime.

[siddhesh0007]

No need for any theorem puys


2^999/2^2*5^2

2^997/5^2 Cancelling 2^2



2^10%25=-1
Thus 2^990%25=-1
2^7%25=3
Thus 2^997%25=-3 &
2^999%25=-12


Thus the remainder is 88

[anil_sharma]

Originally Posted by reachmonil

But, I am a little amazed at why we can't get the answer using the same Inverse formula without dividing by 4. Does that mean, the 2 numbers need to be co-prime? Nay, I don't thin so.

The reason why answer will not be obtained without multiplying by 4 is that we are not calculating the value of fraction but we are calculating value of remainder.
For example

100/75 is same as 4/3
But remainder when 100 is divided by 75 will not be equal to the remainder when 4 is divided by 3. Dividing it by 25, remainder also becomes 1/25 times..


regarding the last 2 digits of the number 2^999, we can use a very simple concept that 76^n always ends with 76 ( as in case of 25)

2^999 = [{2^10}^98][2^19] = (1024)^98. { 2^19}
By taking only last two digits of the product we get,
[24^98][2^19] = [76^49].[ 2^10].[2^9] =76*24*12 = 88