FOURTH GRADE / MATHEMATICS / ADDITION of NATURAL NUMBERS 1.1
MATHEMATICS IS EASY. PROVIDED THAT ONE WORKS ENOUGH .
ENOUGH IS SKILL DEPENDENT, HAVE TO FIND "ENOUGH" RIGHT FOR YOU.
Digits are not counted as letters, make use of the yellow digits.
There are 6 words with a total of 29 letters before the red "k".
There are 6 letters before the red "k" within the same word.
As you see in ry1, the two sentences have some very important properties in common.
Counting forward without taking care of these commons is waste of time.
Both sentences have 82 letters.
7+8 = 15 and the sum has 6 at tens. This implies A+B = 12 .
A can not be 0 , 1 and 2. Otherwise it is not posible to satisfy A+B = 12 .
A can not be 6 , otherwisw B is forced to be also 6.
3 solutions are possible:
1st is to find the sum 1+2+3+4+5 = ... directly.
When the number of consequents increases to 10 000, addition is hard to perform, better drive a formula.
The formula was given in :"http://www.test-dr.com/education/4th-Grade-Mathematics/fourth-grade-natural-numbers-1.2.html" Question 1.
For the geometrical way look at the figure carefully:
The sum of first five integers is the colored region.
The big square has 5x(5+1) small squares .
The no. of white and colored areas are equal.
An equation valid for the natural number 5 should nor necessaryly be valid for the natural number 6.
If there is an implication then we look for the general case:.
Sum the least 5 and 6-digit number and count backward by 2 !
Sum of all numbers between 1 and 10 = 25 x 51
Sum of multiples of 5 between 1 and 50 = 5+10+...+45+50 = 5x(1+2+...+9+10) = 5 x 10/2 x 11
5, 15 , ... 50,51, ..., 55 , ... , 65
One possible solution is to subtract the sum of even numbers from tha sum of all.
Or else we may make all elemnets of the sequence even by subtracting 1 from each and considering this subtraction within the equation somehow.
333/6 = 55.5 implies that the middle two terms should be 55 and 56.
A=4 is a must!
Square of a number is the number multiplied by itself.
Examples : square of 1 = 1x1 = 1 ; square of 5 = 5x5 = 25
Take care that their difference goes as 2,3,4,5, ... .
All of them seem to be true .
The solution of the previous question solves this q. also .
If we had different choices it would help to know: The 15th triangular number is the sum of no. from 1 to 15, i.e. 15 x 16/2.
They are the triangular numbers.
All numbers of PT are integers ...
Use the followings :
Addition of an even and an odd numbers gives an odd number .
If the digit-sum of the summands is a multiple of 3, the digit-sum of the sum is also ...
Remember the def. of factorial and make use of it .
This is the feedback!