Algebra Word Problem/ Riddle. Can you help me out please?

Question

Using all digits 0 through 9 only once, create multiples of 4 that will sum to the smallest number possible. An example is shown below:1472 + 956 + 308 = 2736The three numbers are multiples of 4 and use the digits 0 through 9 once and only once. (However, the sum of 2736 is not the smallest number that is possible.)After solving the problem, you must write a brief paragraph that explains how you solved the problem. This should include specifics on how you determined the addends.

Answer 1

First of all, for a number to be divisible by 4, the last two digits must be a multiple of 4 such as 04, 08, 12, 16, 20, 24, 28, 32, 36, 40, 48, 52, 56, 60, 64, 68, 72, 76, 80, 84, 92, 96...

Notice that the numbers above only end in 2, 4, 6, or 8. Any of our odd numbers will need to be in the tens place or higher. In order to keep our total sum low, we should try to use as many one and two digit numbers as possible.

Let's start with 8. It is a multiple of 4 and a single digit number, that we can use up right away. Since 8 is a larger number, we are best to use it in the ones place. Next, think about 9. As we saw, we cannot make a multiple of 4 with a 9 in the ones place, so we want to use it in the tens place, because 90 is much smaller than 900 or 9000. So we can use either 92 or 96. Most likely, we are better off using 96, and trying to save the 2 as a tens place digit later.

Therefore, so far we are thinking 8 and 96, which totals 104. We have used the two largest numbers we possible could in the ones place (6 & 8) and the 9 in the tens place. These were the smallest place values we possibly could fit them into. The same thinking holds for 7. We want to try to get 7 into the tens place, since we can't get it into the ones place. The only option we have is 72, since we used the 6 in 96. We could also reverse those (76 & 92 instead of 72 & 96) because it wouldn't change our sum.

So far, we have 8 + 96 + 72, for a total of 176. We have yet to use 1, 3, 4, 5 and 0. We are going to have to use 1, 3, and 5 in the hundreds place, but at least they are the smallest three hundreds numbers possible. Add these to the front of our previous three numbers, 8, 96, and 72, in any order you want. Notice that we will use the 0 along with the 8 to make our hundreds numbers.

One possible combination of three numbers is 108, 396, and 572. You can switch around the 1, 3, and 5 anyway you wish, as long as you keep 08, 96 and 72 as the last two of each so they will be divisible by 4. Remember you can also swap 96 and 72 for 92 and 76.

Anyway you arrange them, your total should be:
108 + 392 = 500 + 576 = 1076

Hopefully, this is the lowest possible, but it may not be. Just seems to make the most sense to me to arrange that way!

Answer 2

1236+236+348 = 1820

Answer 3

1048(262 x 4) + 792(198 x 4) + 356(89 x 4) = 2196

These are all multiples of 4. I just used trail and error + 1234/4 = 308.5 so that can't work so I just tried different things.

p.s. The first person answers are not multiples of 4.

Hope this helps
Berethor

if you want me to explain it more send me a email at
[e-mail]

Answer 4

0147+258+369=774

if the 0 must not be on the left
1047+258+369=1674

u should distibute the low numbers in the heigh rank places
for thousands you should put 0 (or 1 if zero shouldn't be in this place)
for handreds distribute the 1,2 and three on the three numbers
for tens distribute the 4,5 and 6 on the three numbers
for tens distribute the 7,8 and 9 on the three numbers

Answer 5

Have a search on google for magic squares and that should help you solve the riddle. Good luck.