Number System Shortcuts, Practice Problems and Solutions - REQMAT BLOGSPOT
Number System Shortcuts, Practice Problems and Solutions

Number System Shortcuts, Practice Problems and Solutions

REQMAT BLOGSPOT - Nareddula Rajeev Reddy NRR

Source (Internet): IACE Material.

Basic Rules On Natural Numbers

1.One digit numbers are from 1 to 9.There are 9 one digit numbers. i.e. 9*100.
2.Two digit numbers are from 10 to 99.There are 90 two digit numbers. i.e. 9*101.
3.Three digit numbers are from 100 to 999.There are 900 three digit numbers. i.e. 9*102.

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. 12+22+32+...+n2=n(n+1)(2n+1)/6.
6.Sum of the cubes of the first n natural numbers i.e. 13+23+33+...+n3=[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 Divisibility:.

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

Example:3582,460,352...…..

Divisibility 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.

Divisibility 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.

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

Example:10,25,60.

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

Example: 48,24,108.

Divisibility 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.

Divisibility 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 .

Divisibility 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 divisible by 9 so, 684 is also divisible by 9.

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

Example: 20,180,350...…

Divisibility 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 (551+552+553+554+555)?
solution : (551+552+553+554+555)
                 551(1+5+25+125+625).
                 551(781).
                 Since 781 is divisible by 11,551*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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