Directions: Using the digits 1 to 9 exactly one time each, place a digit in each box to make the sum as close to 1000 as possible.

### Hint

### Hint

How do you know you can’t get any closer to 1000? What should be true about the hundreds places of your three numbers? How do the tens places affect your answer?

### Answer

### Answer

Lots of answers will get you 999. One would be 247 + 563 + 189.

Is exactly 1000 even possible? Here’s a Geogebra tool you can use to check your answer.

Is exactly 1000 even possible? Here’s a Geogebra tool you can use to check your answer.

Source: John Ulbright and Robert Kaplinsky

One would be 189+247+563

Hi, I’m a 1st time visitor. We are just really getting started recording various levels of DOK questions within our common lessons. I found your site via a Goggle search. However, I have a question. How can I confirm if a question is a DOK 2 or 3? They do not seem to be identified. Are the levels noted somewhere?

Thanks,

Laura

Hi Laura. You can see it in the gray tags area just above the comments section. This problem is DOK 3.

600+300+100=1000

Devn, that isn’t true. You cannot use 0s.

176+534+289=999

152+368+479=999

583+167+249=999

What a rich problem! There are a number of interesting extensions. Without using an exhaustive search, how can we prove it’s not possible to get to exactly 1000?

First observe that we must split or partition the 9 digits into three groups of three digits: hundreds, tens and ones. For any partition, the sum is the same no matter which hundreds digit goes with which tens digit and which ones digit. If the ones place digits are (1, 2, 3) then they will add up to 6 regardless of what is in the tens place and hundreds place for each number. We could switch around the ones place digits and it would not affect the total. The same thing is true for the tens and hundreds place digits. So we can simplify the problem to asking whether there is a partition that adds up to 1000.

To make exactly 1000 without going over, we would need for the ones digits to sum to 20, the tens digits to sum to 18, and the hundreds digits to sum to 8. That way, with the carries, the sum is exactly 1000. A bit of experimentation should convince you that only these sums in each place could work. The hundreds digits can’t sum to 9 because then the tens digits would need to sum to 8 and the ones would need to sum to 20 and there would be too many units

But 20+18+8 =46 and 1+2+3+4+5+6+7+8+9 =45. We don’t have enough units in the digits available to get a sum of digits of 46 which is necessary for the sum of the three digit numbers to be 1000. We will be one unit short in either the hundreds, tens or ones place. 999 is the closest possible result and to get there we want the hundreds digits to sum to 8, the tens digits to sum to 18, and the ones digits to sum to 19.

Another good question: how many possible combinations get us to 999? We can approach this in two steps. One, how many different partitions of the 9 digits will result in sums of 999? Two, for a particular partition, how many different sets of three digit numbers can be made?

We can answer the second part with combinatorics. We can choose from 3 options 3 times for the first number, giving 3^3 possible numbers. Then there are 2^3 possibilities for the second and 1^3 for the third. But since we don’t care about order, and there are 3*2*1 different orders of 3 numbers, overall a particular partition of the numbers yields (3^3 * 2^3 * 1^3) / (3*2*1) = 36 combinations of 3 numbers for a given partition of the 9 digits.

We could get this same value of 36 combinations per partition by thinking of the problem as 3 permutations of 3 digits each divided by the number of ways to arrange 3 objects:, (3 nPr 3) * (3 nPr 3) * (3 nPr 3), divided by (3 nPr 3) . That’s (6*6*6)/6 =36.

Now back to the first step: how many partitions of 9 digits make sums of 999 by making sums of 8 for the hundreds place, 18 for the tens place and 19 for the ones place? I can’t see any other approach but enumeration. There are only two possibilities for the hundreds place: (1, 2, 5) and (1, 3, 4).

If we take (1, 2, 5) for the hundreds place then there are three options for the tens place that sum to 18: (3, 6, 9), (3, 7, 8), (4, 6, 8). If we take (1, 3, 4) for the hundreds place then there are two options for the tens place that sum to 18: (2, 7, 9), (5, 6, 7). Three partitions on the first option plus two partitions on the second optiobakes 5, times 36 combinations per partition makes 180 solutions.

How hard would it be to guess on this problem? The number of solutions is 5*36 =180. The total number of three digit number sets possible is 9!/3!=60480. So 180/60480 =1/336 of the possible guesses is a solution. If you just guessed randomly and checked, you might expect to guess 336 times before getting a lucky answer.

Shout out to Mr. Barton’s Maths Podcast, which brought me here!

To pique student interest and involvement, I put a leaderboard on the classroom whiteboard. This showed first, second, third and fourth places of who in the class could get closest to 1000. It was great to see so many students taking up the challenge to get first place.

By the way, where do I find the DOK that goes with this resource? I’m having trouble finding it on the website.