A Tribute To My First Problem
3 2 0 1 7 − 3 2 0 1 6 = ?
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
6 solutions
Yes, this is indeed true. But because this problem is for newbies, it might be worthwhile to explain why the first and the second expressions are indeed equal.
Log in to reply
Added the required step. I hope now it will be clear.
Wait. a^m-a^n=a^m-n so
Log in to reply
a n a m = a m a − n = a m − n
Kindly check Rules of Exponents
Relevant wiki: Rules of Exponents
3 2 0 1 7 = 3 2 0 1 6 × 3 . Thus, 3 2 0 1 6 × 3 − 3 2 0 1 6 × 1 = 3 2 0 1 6 × 2
I can't understand where is the 2 come from
Log in to reply
If you factor out the 3 2 0 1 6 from the second equation, you get 3 2 0 1 6 × ( 3 − 1 ) which simplifies to the boxed expression.
by taking 3^2016 common.
care to explain how?
Log in to reply
What do you not understand? All we have to know is a b + c = a b × a c .
This works best out of all these explanations for me thanks. I worked it out by calculating 3^3 and 3^4, then seeing that the difference between them was twice 3^3, so thought that it must follow that difference between 3^2017 and 3^2016 must be twice 3^2016
Log in to reply
Yup, playing with small numbers gives us a hint that 3 n − 3 n − 1 is equal to 2 × 3 n − 1 . Then all that remains is substituting n = 2 0 1 7 .
I thought about it differently. The last digit of 3^1 is 3. The last digit of 3^2 is 9. The last digit of 3^3 is 7. And the last digit of 3^4 is 1. This pattern of 3,9,7,1 will repeat. That means that 3^2017 ends in a 3 and 3^2016 ends in a 1. 3-1=2. The answer ends in 2. The last choice is the only possibility.
Log in to reply
Nice observation. The difference of two odd numbers is an even number, so the answer had to be divisible by 2.
Let n=2016
3^(n+1) - 3
3^(n)
3 - 3^(n)
Also let 3^(n)=p
3p - p = 2p
Therefore , 2p = 2
(3^2016)
Yes, the most important step is identifying that a b + 1 = a b × a . Good work!
i did it a lil different, I used differential calc.... 3(^2017)-3(^2016)= 9(^2016)-3(^2016)= 6(^2016). i think anyway id have to pen and paper it..
Multiplication distributes over addition in a field so we have
3 2 0 1 7 − 3 2 0 1 6 = 3 2 0 1 6 ( 3 − 1 ) = 3 2 0 1 6 × 2 .
Yup, the distributive property a ( b + c ) = a b + a c holds true for all real numbers a , b , c .
As a hint to the solution, remember that all integer powers of 3 ( 3 x ) must result in an odd number :
3 2 = 9
3 3 = 2 7
3 4 = 8 1
Additionally, an odd number minus another odd number must yield an even number :
5 − 3 = 1
9 − 5 = 4
From this, the first odd number ( 3 2 0 1 7 ) minus the second odd number ( 3 2 0 1 6 ) yields an even number .
Therefore, the solution must be either one of the even choices: 3 1 × 2 or 3 2 0 1 6 × 2 .
There is a typo in your answer. 5 - 3 is not = 1
Yes, we can eliminate two choices using parity arguments. How did you decide between 3 1 × 2 and 3 2 0 1 6 × 2 ?
Relevant wiki: Rules of Exponents
3 2 0 1 7 − 3 2 0 1 6 = 3 2 0 1 6 × 3 1 − 3 2 0 1 6 = 3 2 0 1 6 ( 3 1 − 1 ) = 3 2 0 1 6 × 2