stne wrote:
EgmatQuantExpert wrote:
Interesting Applications of Remainders – Practice question 2 When \(8^{48}\) is divided by 3 and 5 respectively, the remainders are \(R_1\) and \(R_2\). What is the value of \(R_1\) + \(R_2\)?
To solve question 3: Question 3To read the article: Interesting Applications of Remainders hi
chetan2u,
Can you explain the first part of this question in a better way, how \(\frac{8^{48}}{3}\) gives a remainder of 1?
I understand cyclicity of 8 is 4.
I understand \({8^{48}}\) gives a unit digit of 6 , but 6 divided by 3 gives no remainder , so what am I missing here ?
Hi...
We are looking for remainders when \(8^{48}\) is divided by 3 and 5..
1) Remainder when \(8^{48}\) is divided by 5
As you said 8 has cylicity of 4 and here the remainder will depend on the units digit..
So units digit are 8,4,2,6... As 48 is 4*12, so units digit will be same as that of 8^4..
Hence units digit will be 6 and remainder will be 1 since 6 divided by 5 gives us 1 as remainder.
2) Remainder when \(8^{48}\) is divided by 5
Here we will use the property of divisibility by 3 : If the sum of digits of number is divisible by 3, the number itself is divisible by 3.
So 8 will leave 2 as remainder...
8^2 or 64 should leave 2^2 or 4, which is equal to 1, as remainder.. 64=3*21+1
Next 8^3 will give 2^3 or 8, which is same as 2...
So the remainders have a cylicity of 2,1,2,1,2,1....
Odd power will give 2 and even power give 1 as remainder.
Here 48 is power and hence even. Therefore, remainder will be 1..
Of course other ways are to convert them in binomial expansion..
8^(48)=(9-1)^48...
When you expand this, all terms will have 9 except last term..
Expansion : 9^48+9^47*(-1)^1+9^46*(-1)^2....+9^(-1)^47+(-1)^48...
Only (-1)^48 is not divisible by 3, remainder = 0+0+0+0...+0+1=1
But you can always find your answer with out use of binomial expansion as GMAT does not test or rely on binomial expansion.
Combined remainder = 1+1=2...
B
Thank you, but for the second point , remainder when divided by 3 , why can't we use the unit digit concept?
When finding the remainder while dividing by 5 we use the unit digit concept but don't use this concept when dividing by three, why? How do we decide when to use unit digit concept and when not to use unit digit concept?