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

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)

3 digit problem has to be solved

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

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

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,

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