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This is attributed to an early school lesson when the teacher thought he would keep the class busy whilst he popped out for something. He set the test of adding all the whole numbers from 1-100. By the time he reached the door, Gauss had the answer.
Gauss imagined the problem as 1 + 2 + 3 +........+98 + 99 + 100, but then he wrote the numbers underneath but in reverse order. 100 +99 + 98..........+3 + 2 + 1
So each 100 pairs of vertical numbers added up to 101 so the total was 10100 but this is twice the true answer as each number is included twice. The total is therefore 5050.
This lead to the general formula that the sum of consecutive numbers from 1 to n is n(n+ 1) ÷ 2.
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What are the seven consecutive composite numbers from one to hundred?
We're not sure exactly what the question is looking for, so I made upmy own, and it leads to a fascinating investigation. I hope this is actuallywhat the questioner has in mind.The question seems to be asking: "Find seven consecutive numbers that areall composite."I did a quick investigation, and found some neat stuff. I sure hope my method was valid.-- There actually is a string of seven composite numbers less than 100.They are: 90, 91, 92, 93, 94, 95, 96.-- There are seven strings of five consecutive composite numbers, all less than 100.-- The next series of seven consecutive composites doesn't occur until 360-366.But on the way there, there are five strings of 9 composites, two strings of 11 composites,and three strings of 13 composites.-- There are no series of an even number of consecutive composites, at leastnot up to 200,000 . Wonder why that is.-- The string of 19 consecutive numbers from 888 to 906 are all composite.-- Up to 200,000, the grand-daddy is the range from 155,922 to 156,006 ...a series of 85 consecutive integers that are all composite.
The sum of three consecutive even integers is 138 what are the integers?
Do-it-in-your-head method: The middle number MUST be a third of the total so 138 divided by 3 is 46, making your three numbers 44, 46 and 48. Shazam!
What two consecutive numbers have a product of 1406?
Assuming you mean consecutive numbers are integers, then: The integers will be the two integers between which the square root of 1406 lies. Calculating the square root of 1406 using the "long" division method: ________3__7 _____---------- ___3|_14_06 ________9 _______--- __67|__506 ________469 ________---- _________37 The square root of 1406 is 37.<something>, thus it lies between 37 and 38, so: 1406 = 37 × 38
Example of rule method?
The rule method is used to describe any set of numbers, so put any sequence of numbers in brackets and there you go.
What is the LCM of 11 and 12 and how do you solve it?
11 is prime, so it won't have any factors in common with 12. In this case multiply them and get 132, the LCM.There are many methods of finding the LCM of two numbers.Method 1:Method of prime factorization is one of the methods.Prime factorization of 11 = 11(11 is prime)Prime factorization of 12 = 2x2x3Is there anything common in the factorization of both numbers? Nothing, so in this case we multiply 11 and 12 to get our LCM: 11 x 12 =132.Method 2:There is a relation between LCM and GCF of two numbers a and b:LCM(a,b) x GCF(a,b) = Product of a and bSince 11 and 12 are consecutive numbers then their GCF is 1 and product of 11 and 12 is 132.Putting the values in the relation we get:LCM(11,12) x 1 = 132LCM(11,12) = 132/1 = 132Not every time we need to go through these methods, here is a trick to remember: LCM of two consecutive numbers is equal to their product.11 and 12 are consecutive numbers so their LCM is 11 x 12 = 132.
