MathDB

1

Part of 2008 May Olympiad

Problems(2)

Sum(powers of 2)

Source: May Olimpiad(Olimpiada de Mayo) 2008

2/28/2018
In a blackboard, it's written the following expression
1−2−22−23−24−25−26−27−28−29−210 1-2-2^2-2^3-2^4-2^5-2^6-2^7-2^8-2^9-2^{10}
We put parenthesis by different ways and then we calculate the result. For example:
1−2−(22−23)−24−(25−26−27)−28−(29−210)=403 1-2-\left(2^2-2^3\right)-2^4-\left(2^5-2^6-2^7\right)-2^8-\left( 2^9-2^{10}\right)= 403 and
1−(2−22(−23−24)−(25−26−27))−(28−29)−210=−933 1-\left(2-2^2 \left(-2^3-2^4 \right)-\left(2^5-2^6-2^7\right)\right)- \left(2^8- 2^9 \right)-2^{10}= -933
How many different results can we obtain?
combinatoricsalgorithm
6-digit multiple of 45

Source: XIV May Olympiad (Olimpiada de Mayo) 2008 L1 P1

9/22/2022
How many different numbers with 66 digits and multiples of 4545 can be written by adding one digit to the left and one to the right of 20082008?
number theorymultipleDigits