Find the three last digits of 1 4 3 9 9 9 9 9 9 . What is their sum?
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It would be better to show the working for Extended Euclidean Algorithm: 1 = 7 ( 1 4 3 ) − 1 ( 1 0 0 0 ) .
Euler Totient Function: ϕ ( 1 0 0 0 ) = 1 0 0 0 × ( 1 − 2 1 ) ( 1 − 5 1 ) = 4 0 0
Consider modulo 1 0 0 0
1 4 3 9 9 9 9 9 9 ≡ ≡ ≡ ≡ 1 4 3 1 0 6 − 1 1 4 3 1 0 6 ( m o d 4 0 0 ) ⋅ 1 4 3 − 1 1 4 3 0 ⋅ 1 4 3 1 1 4 3 1
So we want to to find modulo inverse of 1 4 3 modulo 1 0 0 0 , which is easily computed to be 7 because 1 = 7 ( 1 4 3 ) − 1 ( 1 0 0 0 ) .
It would be better to show the working for Extended Euclidean Algorithm: 1 = 7 ( 1 4 3 ) − 1 ( 1 0 0 0 ) .
I got the answer using binomial expansion of (140+3)^999999
This solution has been marked incomplete. What is your to split them up as 1 4 3 = 1 4 0 + 3 ? Why don't we choose 1 4 3 = 1 4 4 − 1 , 1 4 3 = 1 4 1 + 2 , or 1 4 3 = 2 0 0 + 1 4 3 ?
Can you elaborate your solution?
yes offcourse sir but i do not know to type the expressions can you help me
Well, you can start by writing the solution. There are the moderators of this community who can and will help you in improving the presentation.
ya sure! if u have google chrome , then use equation editor (app) from the chrome webstore. its a great help!
We can let n be the exponent of 143. Then we must look for a pattern. If the problem asks us to find what is the last digit or the last two digits, it would be much easier. Let n = 1
143^1 = 143.
n = 2, 143^2 = 20449. But we don't need to find out the other digits.
After solving for the last three digits in 143^n, we multiply that number to 143 to get 143^(n+1). If n = 2, the last three digits are 449. n = 3: 449 x 143 = ...207. Just solve for the last three digits to avoid wasting time. Repeat until you find a pattern. After n = 20, the last three digits repeat.
We divide 999999 by 20, then we get a remainder of 19. if n = 19, the last three digits are 007. Hence, the answer is 7.
Of course there are more elegant ways to solve this but for pattern-oriented people such as me this solution works best.
This solution is very impractical. It's easy to mess up on the calculation. And you actually got lucky that the cycle is a mere 2 0 , generally, the cycle when determining the last 3 digits is at most ϕ ( 1 0 0 0 ) = 4 0 0 . Replacing 1 4 3 by almost any other number gives a cycle greater than 2 0 , which will involve more calculation!
last digit of 143^999999 is same as the last digit of 3^999999......... now last digit of 3^1=3 , 3^2=9 , 3^3=7 , 3^4=1 , 3^5=3 , so after power 4 the repetition start so, now 999999 % 4 = 3 , so answer is 7
This solution has been marked wrong. Like Pi Han Goh has mentioned, your working only shows the last digit of the number.
You have only shown the last digit is 7 , you need to show that the last THREE digits is 0 0 7
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First, we note that finding the last three digits is the same as performing the operation of given number modulo 1 0 0 0 . Also, g cd ( 1 4 3 , 1 0 0 0 ) = 1 . Next, we use the Euler's Theorem (which is a generalization of Fermat's Little Theorem ) to ease out computations of residue by reducing exponents on the given number. The theorem states that:
a ϕ ( n ) ≡ 1 ( m o d n ) ∀ a , n ∈ Z + ∧ g cd ( a , n ) = 1
Note: ϕ ( n ) is the Euler's Totient Function which counts the totatives of n .
ϕ ( 1 0 0 0 ) = ϕ ( 5 3 × 2 3 ) = 1 0 0 0 ( 1 − 2 1 ) ( 1 − 5 1 ) = 4 0 0
∴ 1 4 3 4 0 0 ≡ 1 ( m o d 1 0 0 0 ) ⟹ 1 4 3 1 0 0 0 0 0 0 = ( 1 4 3 4 0 0 ) 2 5 0 0 ≡ 1 ( m o d 1 0 0 0 )
1 4 3 9 9 9 9 9 9 = 1 4 3 1 0 0 0 0 0 0 × 1 4 3 − 1 ≡ 1 × 1 4 3 − 1 ≡ 1 4 3 − 1 ( m o d 1 0 0 0 )
Let z ≡ 1 4 3 − 1 ( m o d 1 0 0 0 ) . Then, we have, 1 4 3 z ≡ 1 ( m o d 1 0 0 0 ) . Using Extended Euclidean Algorithm , we have,
1 4 3 z ≡ 1 ( m o d 1 0 0 0 ) ⟹ z ≡ 7 ( m o d 1 0 0 0 )
∴ 1 4 3 9 9 9 9 9 9 ≡ 7 ( m o d 1 0 0 0 )
Required sum = 0 + 0 + 7 = 7