Latest Elitmus Aptitude Question SOLUTION: 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?

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(#M40013761) ELITMUS Numerical Ability question elit Keep an EYE

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?

Asked In Elitmus Ashiwani Kumar (11 years ago)

Unsolved Read Solution (8)

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Vinay Sipani   (11 years ago)

18x = y

All the digits of y are 4s
18 = 9*2

=> to be divisible by 9,sum of all digits of y is divisible by 9
=> (4 + 4 + 4 + ...) is divisible by 9

LCM of 9 and 4 is 36

so there must be zero or nine 4s to be divisble by 18

=> 18x = (444...9 times) or 0
=> x = 0 or 24691358

Is this solution Helpfull? Yes (44) | No (12)  

CHANDRAMAULI SRIVASTAVA  (11 years ago)

ans- 8
just take a sequence of 4s ,lets start from two 4s and divide it by 18 and keep
on adding number 4 in dividend and keep on dividing it till you get 0 as remainder.Count the digits of the quotient i.e. the answer.

Is this solution Helpfull? Yes (22) | No (7)  

prayas goel  (10 years ago)

Its simple man..
Digit x must be divisible by 18...i.e. must be divisible by 9 also..
and only one condition will satisfy that its must all 4s and divided by 9
i.e. 9 times 4.
444444444/18= 24691358..minimum no. of digits in 24691358 = 8 digits

Is this solution Helpfull? Yes (18) | No (2)  

CRASH  (10 years ago)

18x = 444...444
x = 444...444/18
x = 222...222/9
to be able to be divide by 9 (2+2+2+....n times)= 9x
2n = 9x
minimun x to achieve such condition = 2
n = 9

Is this solution Helpfull? Yes (4) | No (2)  

Mukesh Kr Singh  (11 years ago)

Hello vinay sipani.. x can not be zero.
Because if we take x=0 then y will be also 0..
But it mention thay y should be multiple of 4

Is this solution Helpfull? Yes (2) | No (4)  

Mukesh Kr Singh  (11 years ago)

sorry not multiple its all digit as 4s...

Is this solution Helpfull? Yes (1) | No (4)  

Deepa Gusain  (6 years ago)

18x=y

here y is 44.....4
x can't be zero since y is not zero.
now,

18x=44....4
=> x=44...4/18
=>x=22...2/9

to be divisible by 9, sum of digits of that no must be divisible by 9.
so x=22...2(n times)/9 where n>0
now for n=1 sum =2 , not divisible by 9
for n=2 sum=2+2= 4 or 2*2= 4 ,not divisible by 9
for n=3 sum= 2+2+2=6 or 2*3= 6 ,not divisible by 9
.
.
.
for n=9 sum= 2*9= 18 ie., divisible by 9

so, x=222222222/9
x= 24691358

Is this solution Helpfull? Yes (1) | No   

Ayush  (10 years ago)

If its 4s, the digit can be either 4 or 8. So, to make it minimum we need 9 4's or we can take 4 8's and 1 4's as then the sum will be 36 , divisible by 9. So, minimum digits can be 5.

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