Latest Elitmus Aptitude Question SOLUTION: If a number consists of each digit of each 1 to 9 then the difference of its sums of alternate digits. a)0 b)11 c)22 d)33

As per this question 1 more condition was there that the number is divisible by 11.

Ans : option b that is 11.

explanation :

sum of the digits from 1 to 9 is 45.
and as we know if a number is divisible by 11 then the sum of digits at odd and even places would be zero or multiple of 11.

now as per the question we have x + y = 45
take x as sum of digits at odd or even place of the number.
take y as sum of digits at odd or even place of the number.

so we get x = 45 - y.
as per the given options we have :
x - y = 0 or x- y = 11 or x -y = 22 or x -y = 33 we can get to know which is satisfying the equation.

so as a whole we get 45 - y - y = 0 as we put the value of x from the equation x + y = 45.

so out of 4 only 2 options are giving the integer value. ie 45-2y=11 and 45-2y=33.

=>45-2y=33
=>2y=78
=>y=39.
now x+y =45 therefore x =6.
as x = 6. we cant get 6 as the sum of the digits at even or odd places. therefore it cannot be possible .

next is 45 -2y= 11.
=>2y = 56
=>y=28.
therefore x = 45-28
=> x= 17.
Now we can get 17 as a sum of digits = 17 at odd places and 28 at even places. see the arrangement as 2+9+3+8+5+7+6+1 .

and the difference of 28 and 17 is 11 that is divisible by 11 hence the answer is 11.

Ans:- a) 0

Solution:-

1+2+3+4+5+6+7+8+9 = 45
sum of alternate digit
9+8+7+6+5+4+3+2+1 = 45

then difference
=45-45
=0

there is 1 more condition that number is divisible by 11