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find all the integers which are equal to eleven times the sum of their digits

find all the integers which are equal to eleven times the sum of their digits


 

Grade:12

4 Answers

siddesh mahadevan
4 Points
9 years ago

in my opinion, its not possible for two digit number..

lets say its a two digit number, made of two +ve digits, x & y

10x+y=11(x+y)

so, 10y= -(x)... contradicts!!

3 digit problem has to be solved

100x+10y+z=11(x+y+z)

Sathya
35 Points
9 years ago

But it is possible for 3 digit no 

 

If its a two digit number, made of two +ve digits, x & y

10x+y=11(x+y) which aint possible,

 

THus for three digit no,:

100x+10y+z=11(x+y+z)

10z + y=89x

for this expression,The values for x,y and z which suits the equation is 1,9,8 respectively

If that number have 1 digit then there are no solution.

If that number have 2 digits then  there are no solution.


If that number have more than 4 digits then there are no solution.

Because if that number have  digits then :


1000a+100b+10c+d > 11a+11b+11c+11d     But , so there are no solution.

Now we just find that number is 198

Ashwin Muralidharan IIT Madras
290 Points
9 years ago

Hi Vivek,

 

You should firstly identify the number of digits in the integer.

Lets say we have an integer of the form x1x2x3.....xn (which is an integer of n digits)

So we have 10n-1x1 + 10n-2x2 + 10n-3x3 + ...... 10xn-1 + xn = 11*( x1 + x2 +..... + xn).

 

You can very easily check that this equation can hold true only for a 3 digit number (ie n = 3) or for n=1 (where the integer will be 0)

In the case of a three digit number, say abc

we have 100a + 10b + c = 11a + 11b + 11c

or 89a = b + 10c

which can hold true only for a = 1 b = 9 and c = 8.

Hence the other integer will be 198.

 

So there are only two integers " 0 and 198 " for which the above condition will be true.

 

Hope it helps.

Wish you all the best.

 

Regards,

Ashwin (IIT MadraS).

mycroft holmes
272 Points
9 years ago

First its easy to prove that for , we have 

 

That means the solutions to the above problem is to be found only among numbers with less than 4 digits.

 

Now, we know that if we denote the sum of digits by S(n), 9 divides n - S(n). Since n = 11 S(n) this means 9 divides S(n) and hence n.

 

Also 11 divides n and so we have that n is divisible by 99.

 

Now, its easy to verify that n=198 is the unique solution

 

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