Number System Part - 4

Number System Part - 4 

 Rule : Number of Zeroes in an expression

Ex. Find out the number of zeroes in 8 × 15 × 23 × 17 × 25 × 22 ?

Solution : (2^3 ) × (3×5) × (23) × (17) × (5^2 ) × (11×2)

Zeroes are formed by combination of 2 * 5, here number of pairs of (2, 5) is 3 so the numbers of zeros will be three

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Ex. The product of two numbers is 60480 and their HCF is 12 . Find the numbers ?

Solution :Since HCF s 12, so the two numbers will be multiple of their HCF

let the first number is 12P and the second number be 12Q

    ∴12P × 12Q = 60480

    ∴ P × Q = 420

Now pair of numbers whose product is 420 is

( 420, 1 )     ( 210, 2 )     ( 140, 3 )     ( 105, 4 )     ( 60, 7 )     ( 20, 21 )

Out of these ( 210, 2 ) is not prime so neglected

Now the required numbers will be ( 420×12, 1×12 ) ( 140×12, 3×12 ) ( 105×12, 4×12) ( 60 ×12 , 7×12 ) ( 20×12, 21×12 )

( 5040, 12 ) (1680, 36) ( 1260, 48) ( 720, 84 ) ( 240, 252 ) be the required numbers

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Ex. Find the greatest number that will divide 37, 109 and 157 so as to leave the same remainder in each case ?

Solution :Let the remainder be x, then the numbers :

( 37 - x )     ( 109 - x )     ( 157 - x ) must be divisible by the required number.

Also if two numbers are divisible by the certain number then their difference is also divisible by that number

( 109 - x ) - ( 37 - x ) = 72

( 157 - x ) - ( 109 - x ) = 48

( 157 - x ) - ( 37 - x ) = 120

So, the numbers 72, 48, 120 will also be divisible by that number, So HCF of 72, 48, 120 is 24, therefore required number will be 24

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Ex. Find out the number of zeros at the end of products 20×15×16×44×72×95×25

Solution :

Note : Zeroes can be produced by two ways

(1) If there is any zero at the end of any multiplicand.

(2) If 5 or its multiple are multiplied by any even number.

Now 20×15×16×44×72×95×25 =





So total number of zeros are 5