Distinct Sum

A restatement of this problem basically ask us to minimize $10^3 G + 10^2 (D+H) + 10 (B+E+I) + (A+C+F+J)$ .

We clearly eliminate the thousand by letting $G = 0$ .

To minimize the hundred, we must find the smallest possible value for the sum $D+H$ , clearly attained when $D+H=1+2=3$ .

To minimize the ten, we must find the smallest possible value for the sum $B+E+I$ , attained now for $B+E+I=3+4+5=12$ .

And for the unit, we are left with $A+C+F+J=6+7+8+9=30$ .

This generates us $30 + 12 \times 10 + 3 \times 10^2 + 0 \times 10^3 = \boxed{450.}$

Nice!

We need the minimum value of $\overline{A} + \overline{BC} + \overline{DEF} + \overline{GHIJ}$ .

Looking into the equation, we can say than $A$ must be largest, and $G$ must be smallest.

Let's take the thousand's digit $G = 0$

Next is deciding the hundred's digit. We have $2$ hundred's digits: $H$ and $D$ . Since $H$ is next to $G$ , we take $H = 1$ and $D = 2$

Next is deciding the ten's digit. We have $3$ ten's digits: $I$ , $E$ and $B$ . Since $I$ is next to $H$ and $E$ is next to $D$ , we take $I = 3$ , $E = 4$ and $B = 5$

Next is deciding the unit's digit. We have $4$ unit's digits: $J$ , $F$ , $C$ and $A$ . Since $J$ is next to $I$ , $F$ is next to $E$ and $C$ is next to $B$ , we take $J = 6$ , $F = 7$ , $C = 8$ and $A = 9$

We have all the digits now, let's get the answer,

$\overline{A} + \overline{BC} + \overline{DEF} + \overline{GHIJ} = 9 + 58 + 247 + 0136 = 450$

That's the answer!

1000.G + (D+H).100 + (B+E+I).10 + (A+C+F+J)min ---> G = 0, D + H = 1+2 = 3 , B+E+I = 3+4+5 = 12, A+C+F+J = 6+7+8+9 = 30. ----> 1000.0 + 3.100 + 12.10 + 30.1 = 300 + 120 + 30 = 450. Answer : 450

We note that the sum of $\overline{A}+\overline{BC} +\overline{DEF} +\overline{GHIJ}$ is equal to

$1000G+100(D+H)+10(B+E+I)+1(A+C+F+J)$

This is because $G$ is the thousands digit, so its value is 1000 times whatever number it is, etc.

We want to minimize this sum, so it is best to have the smallest digits be put in the largest value placeholders. To do this, we can have $0$ be a thousands digit, $1, 2$ be in the hundreds digit places, $3,4,5$ be in the tens digit places, and $6, 7, 8, 9$ be in the ones digit places. (See or prove why this will minimize the value of the expression.)

Hence, we have $1000\cdot0+100\cdot3+10\cdot12+1\cdot30=300+120+30=\boxed{450}\ \blacksquare$

First, we can rearrange the sum and get: $1000G+100(D+H)+10(B+E+I)+(A+C+F+J)$ If we want to minimize, we have to make the sums multiplied by the higher coefficients as small as possible. This gives us the following values: $G=0$ $D=1,H=2$ $B=3,E=4,I=5$ $A=6,C=7,F=8,J=9$ Thus the total is: $100(1+2)+10(3+4+5)+6+7+8+9=300+120+30=\boxed{450}$

Looking at the sum we have four digits representing 1s (A, C, F and J), three representing 10s (B, E and I), two representing 100s (D and H) and one representing 1000s (G). To minimize the value of the sum of these, we must make the highest number 1s, and then go into 10s, 100s and 1000s the lower the number, i.e. the four 1s are 9, 8, 7 and 6, the 10s are 5, 4 and 3, the 100s are 2, 1 and the 1000s is 0.

ie. A/C/F/J = 9/8/7/6, B/E/I = 5/4/3, D/H = 2/1 and G = 0

This gives us the sum 9 + 8 + 7 + 6 + 50 + 40 + 30 + 200 + 100 + 0000

Which equals 450.

Note that each digit is only used once in this sum. The first thing that we must do is think. When we add these up, we want the thousands digits to be the smallest, hundreds the next to smallest, etc.. But what's nice is that we don't need to assign specific values, we only need to take it all as one lump sum (Remember that nifty little thing called the commutative property?). We will let the sum of all of the units digits be the largest 4 numbers, 9+8+7+6. These add up to 30. Carry the 3, and let the tens be the next biggest, 5+4+3 plus that extra 3. These add up to 15. Carry the 1, and solve the hundreds with the same procedure, 2+1 plus the extra 1, to get a final answer of 450. (Remember, you learned this in 1st grade.).

A = 9 B = 5 C = 8 D = 1 E = 4 F = 6 G = 0 H = 2 I = 3 J = 7 So, The minimum sum = 9 + 58 + 146 + 0237 = 450

For the overall value to be minimum, going by the place value, all the biggest numbers should be in the smallest place values and the smallest in the biggest place values. So 6, 7, 8 and 9 fall in the units place(in any order), 3,4,5 fall in the tens place(in any order), 1,2 fall in the hundreds place(in any order) and 0 in the thousands place. Hence, taking one of the combinations we get 6+47+238+0159 = 450.

For the minimum value you want the 0 to be in the largest place. You rewrite it as A + 10B + C + 100D + 10E...+J. Then you plug in 0 for G, 1 for H, 2 for D, 3, 4, and 5 in the tens place and the rest in the hundreds place. Adding, you get 450