Eleveny mania

Number Theory Level 3

let a number be 'a'.

let its sum of the digits be 'x'.

Condition: Now if I add up 'x' and 'a' and subtract twice the last digit of 'a' from it, then i get a number divisible by 11.

For example: a=132, then x=3+2+1=6

     Now a+x-2(2)=138-4=134

     and 134 is not divisible by 11.

Find the sum of all the numbers satisfying the above condition.

The answer is 4905.

1 solution

Mrigank Krishan
Jan 14, 2015

Simply, as you all leart in earlier classes, we just have to represent nos. in ones, tens, hundreds........

So, for a one digit number:-

For two digits:-

10a+b

for three digits:-

100a+10b+c

and so on........

we'll now apply the condition what i've said on one digit numbers-

a(the number itself)+a(sum of digits)-2a(twice the last digit)

now apply the condition what i've said on two digit numbers:-

10a+b(the number itself)+a+b(sum of digits)-2b(twice the last digit)

=11a+2b-2b=11a

and we'll keep going on and observe that this condition is satisfied only by two digit numbers, hence their sum can be easily finded by ${ S }_{ n }^{ }=\frac { n }{ 2 } \left[ 2a+(n-1)d \right]$ where n=99-10+1=90 and d=1 and a=10

So the sum is:- ${ S }_{ n }^{ }=\frac { 90 }{ 2 } \left[ 2(10)+(90-1)1 \right] = \boxed{4905}$

Eleveny mania

The answer is 4905.

1 solution

0 pending reports