Numbers System Questions Answer


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Numbers System Questions Answer:-

We are jobriya.co.in is one of the largest online exam material provide and Government jobs notification provider in India.Now in this chapter i am going to tell you number system tricks,number system,number system formula,number system questions. here i am providing you short tricks of number system, number system formula with the help of this tricks you can solve number system questions within second. after all number system tricks and example we also provide number system quiz candidate can take quiz without registration. quiz contain 50 question and total time will be 30 minutes.

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Types of Number

  • Natural Numbers : Counting numbers 1, 2, 3, 4, 5,….. are called natural numbers.
  • Whole Numbers : All counting numbers together with zero form the set of whole numbers. Thus,
    (i) 0 is the only whole number which is not a natural number.
    (ii) Every natural number is a whole number.
  • Integers : All natural numbers, 0 and negatives of counting numbers i.e.,
    {…, – 3 , – 2 , – 1 , 0, 1, 2, 3,…..} together form the set of integers.
    (i) Positive Integers : {1, 2, 3, 4, …..} is the set of all positive integers.
    (ii) Negative Integers : {- 1, – 2, – 3,…..} is the set of all negative integers.
    (iii) Non-Positive and Non-Negative Integers : 0 is neither positive nor negative. So, {0, 1, 2, 3,….} represents the set of non-negative integers, while {0, – 1 , – 2 , – 3 , …..} represents the set of non-positive integers.
  • Even Numbers : A number divisible by 2 is called an even number, e.g., 2, 4, 6, 8,10, etc.
  •  Odd Numbers : A number not divisible by 2 is called an odd number. e.g., 1, 3, 5, 7,9, 11, etc.
  • Prime Numbers : A number greater than 1 is called a prime number, if it has exactly two factors, namely 1 and the number itself. Prime numbers upto 100 are : 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97.
  • Composite Numbers : Numbers greater than 1 which are not prime, are known as composite numbers, e.g., 4, 6, 8, 9, 10, 12.
    Note : (i) 1 is neither prime nor composite.
    (ii) 2 is the only even number which is prime.
    (iii) There are 25 prime numbers between 1 and 100.
  • Co-primes : Two numbers a and b are said to be co-primes, if their H.C.F. is 1. e.g., (2, 3), (4, 5), (7, 9), (8, 11), etc. are co-primes,

number system tricks, number system formula:-

Formulas of Number Series:- 

  1. 1 + 2 + 3 + 4 + 5 + … + n = n(n + 1)/2
  2. (12 + 22 + 32 + ….. + n2) = n ( n + 1 ) (2n + 1) / 6
  3. (13 + 23 + 33 + ….. + n3) = (n(n + 1)/ 2)2
  4. Sum of first n odd numbers = n2
  5. Sum of first n even numbers = n (n + 1)
  6. nth term = xr (n -1)
  7. Sum of n terms = x (1 – rn) , here r < 1
    (1 – r)
  8. Sum of n terms = x (rn – 1) , here r > 1
    (r – 1)

Shortcuts for number divisibility check:- 

  1. A number is divisible by 2, if its unit’s digit is any of 0, 2, 4, 6, 8.
  2. A number is divisible by 3, if the sum of its digits is divisible by 3.
  3. A number is divisible by 4, if the number formed by the last two digits is divisible by 4.
  4. A number is divisible by 5, if its unit’s digit is either 0 or 5.
  5. A number is divisible by 6, if it is divisible by both 2 and 3.
  6. A number is divisible by 8, if the number formed by the last three digits of the given number is divisible by 8.
  7. A number is divisible by 9, if the sum of its digits is divisible by 9.
  8. A number is divisible by 10, if it ends with 0.
  9. A number is divisible by 11, if the difference of the sum of its digits at odd places and the sum of its digits at even places, is either 0 or a number divisible by 11.
  10. A number is divisible by 12, if it is divisible by both 4 and 3.
  11. A number is divisible by 14, if it is divisible by 2 as well as 7.
  12. Two numbers are said to be co-primes if their H.C.F. is 1. To find if a number, say y is divisible by x, find m and n such that m * n = x and m and n are co-prime numbers. If y is divisible by both m and n then it is divisible by
  13. Divisibility By 14 : A number is divisible by 14, if it is divisible by 2 as well as 7.
  14. Divisibility By 15 : A number is divisible by 15, if it is divisible by both 3 and 5.
  15.  Divisibility By 16 : A number is divisible by 16, if the number formed by the last 4 digits is divisible by 16.
    Ex.7957536 is divisible by 16, since the number formed by the last four digits is 7536,
    which is divisible by 16.
  16. Divisibility By 24 : A given number is divisible by 24, if it is divisible by both 3 and 8.
  17. Divisibility By 40 : A given number is divisible by 40, if it is divisible by both 5 and 8.
  18. Divisibility By 80 : A given number is divisible by 80, if it is divisible by both 5 and 16.

Multiplication Short Cut Methods:-

1. Multiplication By Distributive Law :

(i) a x (b + c) = a x b + a x c (ii) ax(b-c) = a x b-a x c.
Ex. (i) 567958 x 99999 = 567958 x (100000 – 1)
= 567958 x 100000 – 567958 x 1 = (56795800000 – 567958) = 56795232042. (ii) 978 x
184 + 978 x 816 = 978 x (184 + 816) = 978 x 1000 = 978000.

2. Multiplication of a Number By 5n : Put n zeros to the right of the multiplicand and divide the number so formed by 2n
Ex. 975436 x 625 = 975436 x 54= 9754360000 = 609647600

Mathematical Formulas:- 

  1. (a + b)(a – b) = (a2 – b2)
  2. (a + b)2 = (a2 + b2 + 2ab)
  3. (a – b)2 = (a2 + b2 – 2ab)
  4. (a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca)
  5. (a3 + b3) = (a + b)(a2 – ab + b2)
  6. (a3 – b3) = (a – b)(a2 + ab + b2)
  7. (a3 + b3 + c3 – 3abc) = (a + b + c)(a2 + b2 + c2 – ab – bc – ac)
  8. When a + b + c = 0, then a3 + b3 + c3 = 3abc
  9. (a + b)n = an + (nC1)an-1b + (nC2)an-2b2 + … + (nCn-1)abn-1 + bn

3. Division Algorithm or Euclidean Algorithm:– If we divide a given number by another number, then :
Dividend = (Divisor x Quotient) + Remainder
IX. {i) (xn – an ) is divisible by (x – a) for all values of n.
(ii) (xn – an) is divisible by (x + a) for all even values of n.
(iii) (xn + an) is divisible by (x + a) for all odd values of n.

Solved Example of Number System

Ex.What value will replace the question mark in each of the following equations ?
(i) ? – 1936248 = 1635773 (ii) 8597 – ? = 7429 – 4358
Sol. (i) Let x – 1936248=1635773.Then, x = 1635773 + 1936248=3572021.
(ii) Let 8597 – x = 7429 – 4358.
Then, x = (8597 + 4358) – 7429 = 12955 – 7429 = 5526.

Ex.Simplify : (i) 5793405 x 9999 (ii) 839478 x 625
Sol.
i)5793405×9999=5793405(10000-1)=57934050000-5793405=57928256595.b
ii) 839478 x 625 = 839478 x 54 = 8394780000/16 = 524673750.

Ex. . Evaluate : (i) 986 x 237 + 986 x 863 (ii) 983 x 207 – 983 x 107
Sol.
(i) 986 x 137 + 986 x 863 = 986 x (137 + 863) = 986 x 1000 = 986000.
(ii) 983 x 207 – 983 x 107 = 983 x (207 – 107) = 983 x 100 = 98300.

Ex. . Simplify : (i) 1605 x 1605 ii) 1398 x 1398
Sol (i) 1605 x 1605 = (1605)2 = (1600 + 5)2 = (1600)2 + (5)2 + 2 x 1600 x 5
= 2560000 + 25 + 16000 = 2576025.
(ii) 1398 x 1398 – (1398)2 = (1400 – 2)2= (1400)2 + (2)2 – 2 x 1400 x 2
=1960000 + 4 – 5600 = 1954404.

Ex. 7. Evaluate : (313 x 313 + 287 x 287).

Sol. (a2 + b2) = 1/2 [(a + b)2 + (a- b)2]

(313)2 + (287)2 = 1/2 [(313 + 287)2 + (313 – 287)2] = ½[(600)2 + (26)2]

= 1/2 (360000 + 676) = 180338.

Ex.Which of the following are prime numbers ?
(i) 241 (ii) 337 (Hi) 391 (iv) 571
Sol.
(i) Clearly, 16 > Ö241. Prime numbers less than 16 are 2, 3, 5, 7, 11, 13.
241 is not divisible by any one of them. 241 is a prime number.
(ii) Clearly, 19>Ö337. Prime numbers less than 19 are 2, 3, 5, 7, 11,13,17.
337 is not divisible by any one of them. 337 is a prime number.
(iii) Clearly, 20 > Ö39l”. Prime numbers less than 20 are 2, 3, 5, 7, 11, 13, 17, 19.
We find that 391 is divisible by 17. 391 is not prime.
(iv) Clearly, 24 > Ö57T. Prime numbers less than 24 are 2, 3, 5, 7, 11, 13, 17, 19, 23.
571 is not divisible by any one of them. 571 is a prime number.

Ex. 9. Find the unit’s digit in the product (2467)163 x (341)72.

Sol. Clearly, unit’s digit in the given product = unit’s digit in 7153 x 172.

Now, 74 gives unit digit 1.

7152 gives unit digit 1,

7153 gives unit digit (l x 7) = 7. Also, 172 gives unit digit 1.

Hence, unit’s digit in the product = (7 x 1) = 7.

Ex. Find the total number of prime factors in the expression (4)11x (7)5x (11)2.

Sol. (4)11x (7)5 x (11)2 = (2 x 2)11 x (7)5 x (11)2 = 211 x 211 x75x 112 = 222 x 75 x112

Total number of prime factors = (22 + 5 + 2) = 29.

Ex.12. Simplify : (i) 896 x 896 – 204 x 204

(ii) 387 x 387 + 114 x 114 + 2 x 387 x 114

(iii) 81 X 81 + 68 X 68-2 x 81 X 68.

Sol.

(i) Given exp = (896)2 – (204) 2 = (896 + 204) (896 – 204) = 1100 x 692 = 761200.

(ii) Given exp = (387) 2+ (114) 2+ (2 x 387x 114)

= a2 + b2 + 2ab, where a = 387,b=114

= (a+b) 2 = (387 + 114 ) 2 = (501) 2 = 251001.

(iii) Given exp = (81) 2 + (68) 2 – 2x 81 x 68 = a2 + b2 – 2ab,Where a =81,b=68

= (a-b) 2 = (81 –68) 2 = (13) 2 = 169

 

ExWhich of the following numbers is divisible by 3 ?
(i) 541326 (ii) 5967013
Sol.
(i) Sum of digits in 541326 = (5 + 4 + 1 + 3 + 2 + 6) = 21, which is divisible by 3.

Hence, 541326 is divisible by 3.
(ii) Sum of digits in 5967013 =(5+9 + 6 + 7 + 0+1 +3) = 31, which is not divisible by 3.
Hence, 5967013 is not divisible by 3.

Ex.What least value must be assigned to * so that the number 197*5462 is r 9 ?
Sol.
Let the missing digit be x.
Sum of digits = (1 + 9 + 7 + x + 5 + 4 + 6 +»2) = (34 + x).
For (34 + x) to be divisible by 9, x must be replaced by 2 .
Hence, the digit in place of * must be 2.

Ex. Which of the following numbers is divisible by 4 ?
(i) 67920594 (ii) 618703572
Sol.
(i) The number formed by the last two digits in the given number is 94, which is not divisible by 4.
Hence, 67920594 is not divisible by 4.
(ii) The number formed by the last two digits in the given number is 72, which is divisible by 4.
Hence, 618703572 is divisible by 4.

Ex. Which digits should come in place of * and $ if the number 62684*$ is divisible by both 8 and 5 ?
Sol.
Since the given number is divisible by 5, so 0 or 5 must come in place of $. But, a number ending with 5 is never divisible by 8. So, 0 will replace $.
Now, the number formed by the last three digits is 4*0, which becomes divisible by 8, if * is replaced by 4.
Hence, digits in place of * and $ are 4 and 0 respectively.

EX. Show that 4832718 is divisible by 11.
Sol. (Sum of digits at odd places) – (Sum of digits at even places)
= (8 + 7 + 3 + 4) – (1 + 2 + 8) = 11, which is divisible by 11.
Hence, 4832718 is divisible by 11.

Ex. Is 52563744 divisible by 24 ?
Sol. 24 = 3 x 8, where 3 and 8 are co-primes.
The sum of the digits in the given number is 36, which is divisible by 3. So, the given number is divisible by 3.
The number formed by the last 3 digits of the given number is 744, which is divisible by 8. So, the given number is divisible by 8.
Thus, the given number is divisible by both 3 and 8, where 3 and 8 are co-primes.
So, it is divisible by 3 x 8, i.e., 24.

Ex. What least number must be added to 3000 to obtain a number exactly divisible by 19 ?
Sol. On dividing 3000 by 19, we get 17 as remainder.
Number to be added = (19 – 17) = 2.

Ex. What least number must be subtracted from 2000 to get a number exactly divisible by 17 ?
Sol. On dividing 2000 by 17, we get 11 as remainder.
Required number to be subtracted = 11.

Ex. Find the number which is nearest to 3105 and is exactly divisible by 21.
Sol. On dividing 3105 by 21, we get 18 as remainder.

Number to be added to 3105 = (21 – 18) – 3.
Hence, required number = 3105 + 3 = 3108.

Ex. Find the smallest number of 6 digits which is exactly divisible by 111.
Sol. Smallest number of 6 digits is 100000.
On dividing 100000 by 111, we get 100 as remainder.
Number to be added = (111 – 100) – 11.
Hence, required number = 100011.-

Ex. On dividing 15968 by a certain number, the quotient is 89 and the remainder is 37. Find the divisor.
Dividend – Remainder                                 15968-37
Sol. Divisor = ————————– = ————- = 179.
Quotient                          89

Ex. A number when divided by 342 gives a remainder 47. When the same number ift divided by 19, what would be the remainder ?
Sol. On dividing the given number by 342, let k be the quotient and 47 as remainder.
Then, number – 342k + 47 = (19 x 18k + 19 x 2 + 9) = 19 (18k + 2) + 9.
The given number when divided by 19, gives (18k + 2) as quotient and 9 as remainder.

Ex.  Find the remainder when 231 is divided by 5.
Sol. 210 = 1024. Unit digit of 210 x 210 x 210 is 4 [as 4 x 4 x 4 gives unit digit 4].
Unit digit of 231 is 8.
Now, 8 when divided by 5, gives 3 as remainder.
Hence, 231 when divided by 5, gives 3 as remainder.

Ex. How many numbers between 11 and 90 are divisible by 7 ?
Sol. The required numbers are 14, 21, 28, 35, …. 77, 84.
This is an A.P. with a = 14 and d = (21 – 14) = 7.
Let it contain n terms.
Then, Tn = 84 => a + (n – 1) d = 84
=> 14 + (n – 1) x 7 = 84 or n = 11.
Required number of terms = 11.

 

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