Not 1, not 2 .. but 13 times the digit sum of N

N canot be one-digit. If N= $\overline{ab}$ , (a two digit number), then 13a+13b=10a+b, which is impossible. So N is three digit number. So if N= $\overline{abc}$ , then

$13a+13b+13c=100a+10b+c$ , so that $b+4c=29a$ . so $b+4c$ is divisible by 29 Note that $b+4c \leq 45$ So $b+4c=29$ and $a=1$ So the posible solutions for (b,c) are (1,7),(5,6),(9,5).

So $117+156+195=468$

First, we check if N can be a two-digit number. Let N = 10a + b, wherein a is a one-digit positive integer, and b is a one-digit integer greater than or equal to 0. When we equate (10a + b) with 13(a + b), we find that 3a + 12b = 0 -- which is impossible, unless a and b are both equal to 0.

Next, we check if N can be a three-digit number. Let N = 100a + 10b + c, wherein a is a positive one-digit number, and b and c are one-digit integers greater than or equal to 0. When we equate (100a + 10b + c) with 13(a + b + c), we find that 87a - 3b - 12c = 0. The equation can be simplified to 29a - b - 4c = 0. There are three possible solutions for the equation: (1, 1, 7), (1, 5, 6), and (1, 9, 5). Hence, there are three possible values for N: 117, 156, and 195.

If we add the three possible values, we get 468. Therefore, the answer is 468.

Let the digits in hundredth, tenth and unit place be a,b,c. Then we have to solve the equation 13(a+b+c)=100a+10b+c, which is reduced to 29a-4c=b. If a=1, b=29-4c. Note that a,b,c takes values 0 to 9. So c=5 gives b=9; c=4 gives b=5 and c=7 gives b=1. Thus the numbers are 195,156 and 117. For a=2, we get b=58-4c and for no value of c from 0 to 9 gives b in that range. In fact for all values of a greater than 1, we do not get b and c in the range.so theses are the only 3 numbers satisfying the conditions. so their sum is 195+156+117=468, which the required answer!

Case 1: N is a 2-digit number.

N could be expressed as 10 a + b where a and b could be any integer from 1 to 9 and b could be 0.

From the question. we could say that:

10 a + b = 13 a +13 b

-12 b = 3 a

Since neither a nor b could be negative, N could not be a 2-digit number.

Case 2: N is a 3-digit number. Let b and c be integers ranging from 1 to 9 and a be an integer ranging from 1 to 9. Establishing the condition, we have:

100 a + 10 b + c = 13 a + 13 b + 13 c

87 a = 3 b + 12 c

29 a = b + 4 c

From the conditions, we could see that when

b = 1, c = 7, then a = 1

b = 5, c = 6, then a = 1

b = 9, c = 5, then a = 1

so the numbers that satisfy the question are 117, 156 and 195. Adding the numbers gives us 468.

However, it could be proven that there are no 4-digit number or greater that is equal to 13 times its digits.

There are a maximum of 27 possible integers N because the maximum digit sum of N is 9+9+9 = 27 i.e. N is at most 27*13.

From there, just trial and error and you should get three numbers 117 = 9 13, 156 = 12 13, 195 = 15*13. It is clear that 1 + 1 + 7 = 9, 1 + 5 + 6 = 12, 1 + 9 + 5 = 15

N must be a three digit number as no such number is possible with one or two digits, so let N = abc = 100a + 10b +c = 13(a+b+c).

Therefore 87a = 3b+12c, where 1<=a,b,c<=9.

Looking at multiples of 87, we find that only a=1 is possible and there are three possibilities for b and c, namely 117, 156 and 195.

Sum of these numbers is 468.

N cannot be 1000 since 1000 is not a multiple of 13. So let the 3 digits of N be a,b and c. Then we have: 100a+10b+c = 13a+13b+13c 100a-13a+10b-13b+c-13c = 0 87a-3b-12c = 0 87a = 3b+12c 29a = b+4c a <9, b <9, c_<9 a = 1. If a had been 2 and b had been 9, c = (29x2-9)/4 >9 29 = 1 + 28, 5+24 or 9+20. Hence N = 117, 156 or 195. 117+156+195 = 468.

let the number N be form 100a + 10 b+ c. then as per given condition,

13(a + b+ c) = N 13(a+ b+ c) = 100a + 10 b+c

This implies 3b +12c = 87a

therfeore the maximum value of (b,c) = 9 and on substituting it we get maximum value of 3b + 12c = 135 This implies a has to be equal to 1 as 'a' cannot be 0.

therefore we get 3b + 12c= 87 , by putting a =1,

By trial and error we get the only possible values are (b,c)= (1,7)(5,6)(9,5) Therefore the possible values of N = 117, 156, 195 Therefore 117 +156+195 = 468

Let kji be the desired integer. So, 13(k+j+i)=100k + 10j + i or 29k=4i +3j for 0<= i, j <=9 and 0<k<=9. Also notice that for the number 999 the digit sum takes the maximum value 27 and 27* 13=351. That means that k=1, 2 or 3. For that point it is easy to find out the solutions.

only the numbers 117,156 and 195 can solve the solution.above 200's no number solve this equation.

we have 3 .they are 117 (1+1+7=9;13 9=117),156(1+5+6+=12;13 12=156),195(1+9+5=15;13*15=195) ,sum=117+156+195=468; simple solution is multiples of 39 which follows the given conditions

If $N = \overline{a}$ for $1 \leq a \leq 9$ , then $a = 13a$ has no solutions.

If $N=\overline{ab} = 10a + b$ for $1 \leq a \leq 9$ and $0 \leq b \leq 9$ , then $10a + b = 13 (a+b)$ $\Rightarrow 0=3a+12b$ also has no solutions as $a\geq 1, b\geq 0$ .

If $N=\overline{abc} = 100a + 10b + c$ for $1 \leq a \leq 9, 0 \leq b \leq 9$ and $0 \leq c \leq 9$ , then $100a + 10b + c = 13(a+b+c)$ $\Rightarrow 29a = b+4c$ .

Since $b+4c \leq 9 + 4 \times 9 = 45 < 58 = 29\times 2$ , thus we must have $a =1$ as the only possibility. So we solve $b+4c=29$ . Considering modulo 4 we see that $b \equiv 1\pmod{4}$ , thus $b = 1,5$ or $9$ which gives solutions $(a,b,c)=(1,1,7), (1,5,6),$ and $(1,9,5)$ .

Hence the sum of all solutions is $117+156+195= 468$ .

If we take a three digit number "abc" (a, b and c are integers between 0 and 9)

13 (a+b+c)=100 a+10*b+c

3b+12c=87a

The largest numbers for both b and c is 9

3x9+12x9>=87a

a<=1

If a=0

3b+12c=0

non of the integers can be under zero

a=1

87=3b+12c

29=b+4c

b can't be more then 9. Therefor c>=5

Also, b>=0. So 29-4c>=0, c<8

8>c>=5

a=1 c=5 b=29-4*5=9 195

a=1 c=6 b=5 156

a=1 c=7 b=1 117

195+156+117=468

So this is how I did it. A little process of elimination and some algebra.

We start with the condition that;

$13a+13b+13c=\overline { abc }$

Since the number abc can be expressed algebraically as $100a+10b+c$ we then have

$13a+13b+13c=100a+10b+c$

Rearranging we get

$-87a+3b+12c=0$

If we start with the condition where $a=1$ then we have the equation

$3b+12c=87$

Since b and c can only range from 0 through 9, process of elimination is not difficult here. If $b=9$ , then $c=5$ , so $195$ satisfies the original conditions. If $c=4$ , then $b=13$ , so we know this can't happen. If $c=5$ , then $b=6$ , so $156$ is also a solution. If $c=7$ , then $b=1$ , so $117$ is a solution and this is the lowest number we can get with $a=1$ .

We can check for the condition when $a=2$ . This gives

$3b+12c=174$

If we plug in 9 for both "b" and "c" we get 135 and since $0\le b\le 9$ and $0\le c\le 9$ , then $a\ngeq 2$ .

If we check the condition where $a=0$ , then we get

$3b+12c=0$

Since neither "b" nor "c" can be negative integers, $a\neq 0$ , thus the only solutions are the three we found when $a=1$ , so, the sum of these solutions is

$\boxed { 117+156+195=468 }$

100X+10Y+Z=13 (X+Y+Z).ie 29X=Y+4Z. (X Y Z)=(117),(195),(156) .SUM=468.

digitsSum1 = function(num){ var numAsString = num.toString(); var ones = numAsString.charAt(0); //console.log(ones); digitSum = parseInt(ones); return digitSum; }

digitsSum2 = function(num){ var numAsString = num.toString(); var tens = numAsString.charAt(0); var ones = numAsString.charAt(1); //console.log(tens+" "+ones); digitSum = parseInt(tens)+parseInt(ones); return digitSum; }

digitsSum3 = function(num){ var numAsString = num.toString(); var hundreds = numAsString.charAt(0); var tens = numAsString.charAt(1); var ones = numAsString.charAt(2); //console.log(hundreds+" "+tens+" "+ones); digitSum = parseInt(hundreds)+parseInt(tens)+parseInt(ones); return digitSum; }

digitsSum123 = function(num){ if(num<10){ return digitsSum1(num); } else if(num>=10 && num<100){ return digitsSum2(num); } else { return digitsSum3(num); } }

//console.log(digitsSum123(978)); console.log(" * * * * * * * * * * * "); var sum=0; for(var i=1; i<=1000; i++){ var temp = digitsSum123(i); if(i == temp 13){ console.log(i+":"+temp 13); sum+=i; } } console.log("The sum is "+sum+".");