Question

Find the sum of all 3-digit positive numbers N that satisfy the condition that the digit sum of N is 3 times the digit sum of N+3.

Answer 1

i found 117,120 ; 207,210

Answer 2

to be 2 of them ant more?

Answer 3

Very interesting question.

Answer 4

Suppose you have \[
N =d_1d_2d_3\\
N+3 =d_1d_2d_4 \\
\]Seems like in this case it won't work..

Answer 5

We'll have to take cases maybe .
Let the number be a b c
1st case :
c<=6

then,
a+b+c = 3(a+b+c+3)

2(a+b+c) = -9
so no solution

case 2:
c=7
(a+b+7) = 3( a + (b+1) + 0 )
4 = 2(a+b)
a+b = 2

(a,b,c) can be (1,1,7) and (2,0,7)

case 3:
c=8
(a+b+8) = 3(a+ (b+1) + 1)
2 = 2(a+b)
a+b=1
(a,b,c) can be (1,0,8) only.

case 4:
c=9
(a+b+9) = 3(a+ (b+1) +2 )
3 = 2(a+b)
No solutions

120 and 210 are not solutions! how did you write em? o.O

Answer 6

Only solutions thus are 108,117 and 207 in case I missed something ?

Answer 7

108, 117 and 207 are the only solutions, you didn't miss anything

Answer 8

Nice!!

Answer 9

why did u take \[c \le 6\]

Answer 10

and how did u get a+b+c = 3(a+b+c+3)??

Answer 11

@shubhamsrg ??

Answer 12

for c>6,
the last digit does not remain <9 , hence I took a case for c<6.

and (a+b+c) = 3(a+b+ (c+3)) , that is only what the question says.

Answer 13

\[\huge c=7=>a+b+c=3(a+b+10)\]

Answer 14

\[\huge 2(a+b)=-23\]

Answer 15

@shubhamsrg ??

Answer 16

at c= 7,
on addition of 3, last digit becomes 0 and b gets succeeded by 1.
hence new eqn will be
a+b+ 7 = 3 ( a+ (b+1) + 0 )

Answer 17

\[\huge thanks\]