Question

when a certain number'x' is multiplied by 18, the product 'y' has all digit as 4s. what is the minimum number of digits 'x' can have?

Answer 1

here,we have to find the least number of all digits as 4s divisible by 18 .

now,we know that the no. divisible by borh 2 and 9 is divisible by 18.

Since all the digit is 4,all the possible no. are divisible by 2.So we have to find the least no. of all 4 as digits divisible by 9.

again, we know that if the digitalroot of a number is 0 or divisible by 9,then the no. is divisible by 9.

so, we found that 444444444 is the reqd. no. which have 9 digit.

since,4+4+4+4+4+4+4+4+4=36 divisible by 9.

so,the required number of digit is 9(nine)

HOPE THAT I HELP U SOMETHING

Answer 2

Answer:

Concept :

A positional numeral system uses a single sign, such as "2," or a set of symbols, such as "25," to represent numbers. This single symbol is known as a numerical digit. The term "digit" refers to the ten digits of the hands, which are also known as the decimal (old Latin adjective decem meaning "ten") digits, and which correspond to the ten symbols of the standard base-10 numeral system. The absolute value of the base determines how many different digits are necessary for a given numeral system with an integer basis. For instance, the binary system (base 2) only needs two digits, whereas the decimal system (base 10) requires ten digits (0 through 9) and (0 and 1).

Step-by-step explanation:

Explanation:

$x^{18}$ = 444...

x = (444....)/18

x = (222....)/9

Any number is divisible by 9 only when sum of all its digits is divisible by 9 . So,

2+2+2..9 times =18 which is divisible by 9 . So,

x = 222222222/9 = 24691358, which has 8 digits.

Therefore the minimum number of digits x can have is 8.

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