Source (Internet): IACE Material.

Result(s):

*To get the concept, Number System Shortcuts, Practice Problems and Solutions.

GOfreshers |

**Basic Rules On Natural Numbers**

1.One digit numbers are from 1 to 9.There are 9 one digit numbers. i.e. 9*10

^{0}.
2.Two digit numbers are from 10 to 99.There are 90 two digit numbers. i.e. 9*10

^{1}.
3.Three digit numbers are from 100 to 999.There are 900 three digit numbers. i.e. 9*10

^{2}.

**In general the number of n digit numbers are 9*10**

^{(n-1)}.
4.Sum of the first

*n*natural numbers i.e. 1+2+3+4+...+n=n(n+1)/2
5.Sum of the squares of the first

*n*natural numbers i.e. 1^{2}+2^{2}+3^{2}+...+n^{2}=n(n+1)(2n+1)/6.
6.Sum of the cubes of the first

*n*natural numbers i.e. 1^{3}+2^{3}+3^{3}+...+n^{3}=[n(n+1)/2]^{2}.

**Example:**What is the value of 51+52+53+....+100 ?.

*solution*51+52+53+...+100=(1+3+...+100)-(1+2+3+...+50).

=(100*101/2)-(50*51/2)=5052-1275=3775.

**Rules for Divisiblity :**.

**Divisiblity by 2:**A number is divisible by 2 when the digit at ones place is 0,2,4,6,8.

**Example:**3582,460,352.......................

**Divisiblity by 3:**A number is divisible by 3 when sum of all digits of a number is divisible by 3.

**Example:**453=4+5+3=12.

**Divisiblity by 4:**A number is divisible by 4 if the number formed with its last two digits are divisible by 4.

**Example:**If we take number 45024, the last two digits form 24.Since, the number 24 is divisible by 4, the number 45024 is also divisible by 4.

**Divisiblity by 5:**A number is divisible by 5 if its last digit is 0 or 5.

**Example:**10,25,60.

**Divisiblity by 6:**A number is divisible by 6 if it is divisible by both 2 and 3.

**Example:**48,24,108.

**Divisiblity by 7:**A number is divisible by 7 when the difference between twice the digit at ones place and the number formed by other digits is either 0 or divisible by 7.

**Example:**658 see here 65 - 2*8=65-16=49.

As 49 is divisible by 7 the number 658 is also divisible by 7.

**Divisiblity by 8:**A number is divisible by 8, if the number formed by the last 3 digits of the number is divisible by 8.

**Example:**If we take the number 57832,the last three digits form 832.Since, the number 832 is divisible by 8, the number 57832 is also divisible by 8 .

**Divisiblity by 9:**A number is divisible by 9, if the sum of all the digits are divisible by 9.

**Example:**984=6+8+4=18.

18 is divisibley 9 so, 684 is also divisible by 9.

**Divisiblity by 10:**A number is divisible by 10, if its last digit is 0 .

**Example:**20,180,350............

**Divisiblity by 11:**A number is divisible by 11, when the difference between the sum of its digits in odd places and in even places is either 0 or divisible by 11.

**Example:**30426

3+4+6=13.

0+2=2.

13 - 2 = 11.

As the difference is divisible by 11 the number 30426 is also divisible by 11.

**1)**The sum of three consecutive natural numbers each divisible by 3 is 72.What is the largest among them?

**solution :**3x+(3x+3)+(3x+6)=72.

9x+9=72.

9x = 72-9.

x=63/9=7.

The largest of them is 27.

**2)**How many numbers up to 700 are divisible by both 3 and 5 ?

**solution :**Quotient when 700 is divided by the LCM of 3 and 5 i.e., 15 is 46.

**3)**The sum of the numerator and denominator of a fraction is equal to 5.Five times the numerator is 4 more than twice the denominator . The fraction is :

**solution :**Let the fraction be x/y

x+y=5 and 5x-2y=4.

Solving, x=2 and y=3.

The fraction is 2/3.

**4)**Which of the following numbers will completely divide (5

^{51}+5

^{52}+5

^{53}+5

^{54}+5

^{55})?

**solution :**(5

^{51}+5

^{52}+5

^{53}+5

^{54}+5

^{55})

5

^{51}(1+5+25+125+625).
5

^{51}(781).
Since 781 is divisible by 11,5

^{51}*781 is also divisible by 11.

**5)**Find the sum of all natural numbers from 100 to 175.

**solution :**From 100 to 175 mean including both 100 and 175.

Sum of natural numbers up to 99.

=(99*100)/2=4950.

Sum of natural numbers up to 175.

=(175*176)/2=15400.

Sum of all natural numbers from 100 to 175=15400-4950=10450.

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*To get the concept, Number System Shortcuts, Practice Problems and Solutions.