What is the sum of all even numbers from 1 to 40?
Even numbers up to 40 are 2,4,6,8,10,12,14,16,18,20,22,24,26,28,30,32,34,36,38,40The above series is in A.P. here a=2, d=2, l=40, n=20Sn=n/2[a+l]==>Sn=10[2+40]=420.
How can we select n numbers from a given sequence of number such that gcd of numbers is not equal to 1?
You can’t always make this choice. For instance, consider a set of [math]n[/math] primes. You are missing something here.
What is the sum of all even numbers from 0-1000?
250,500 … There are 501 even numbers if one starts with zero and ends with 1,000. A quick way to determine their sum is to make 501 addition problems starting with 1000 + 0 … 998 + 2 … 996 + 4 … 994 + 6 … etc. all the way to 0 + 1000. Now, you have 501 sums that each add up to 1000. Since the 501 sums are all the same … 1,000 … you can find their total sum through multiplication … 501 x 1,000. That’ll give you 501,000. BUT because you actually added the even numbers from 0 to 1,000 twice … that 501,000 is twice the sum you want. So … you simply divide the 501,000 in half and get 250,500.
What is the minimum number of total moves needed for winning a Solitaire game?
Theoretically, for a very optimal deal of Solitaire it would only take 76 moves for 1-deal, or 60 for 3-deal (one for each card to move into the Foundations and some overflow for dealing from the deck).Imagine a situation with the following deal.There are seven piles, four have A's, three have 2's.After placing the Aces, our next row can have four 2's and two 3's. Continuing in this method, there's a deal where all 28 of the cards below would have been successfully placed in the Foundations.If we are playing 1-deal, there must be a deal where the remaining 24 cards are sorted. So, we get 24 more moves unveiling each card, and 24 more placing each card onto Foundation, giving us a total of 28 + 24 + 24 = 76.If we're playing 3-deal, every group of 3 must be reverse sorted as they come out, but there's still a possible stack where we can simply place cards one at a time increasing the Foundation stack. So, one move unveils 3 cards, so we'll draw from the deck 8 times, and add each card finally onto the Foundation for a total of 28 + 8 + 24 = 60.With either style of play, it's going to be a very lucky deal to get the minimum number of moves!
How can I establish that the sequence (n+1)! -2, (n+1)! -3,…, (n+1)! -(n+1) produces n consecutive composite integers for n>2?
To find those integers are composite, all you need to do is find a divisor of each that’s smaller than itself, yet bigger than [math]1[/math]. As [math](n+1)![/math] contains a factor of [math]2,3,\ldots, n+1[/math], it’s divisible by each of those, so [math](n+1)!-a[/math] is divisible by [math]a[/math] for each [math]2\leq a\leq n+1[/math]. As [math]n>2[/math], we have [math](n+1)!-a>a>1[/math] because [math](n+1)![/math] contains at least one factor bigger than [math]2[/math] that isn’t [math]a[/math], so indeed, the [math](n+1)!-a[/math] form [math]n[/math] consecutive composite integers.