Home > Grade 6 > Dividing Mixed Numbers

# Dividing Mixed Numbers

Directions: Using the digits 1 to 9 at most one time each, fill in the boxes to make the smallest (or largest) quotient.

### Hint

How can we tell whether it would be better to put the larger/smaller numbers as the whole number, numerator, or denominator?

Julie Wright has found the current best answer for largest quotient: 9 8/2 ÷ 1 3/7 and smallest quotient: 1 3/7 ÷ 9 8/2.  See her fantastic write up below. Do you have a better one?

Source: Robert Kaplinsky

## Prime Factorization 2

Directions: Using the digits 0 to 9 at most one time each, fill in the …

1. I found different bigger/smaller one, and I’m pretty sure it’s the optimal pair of mixed numbers. I went down a few blind alleys but like they say, the mistakes kept leading me to learn more about the math. This is the gist:

I wrote the boxes out as variables, at first, so: a b/c divided by d e/f

At first I kind of messed around with algebraic rearrangement, but I wasn’t getting anywhere since the f’s and c’s show up several places. So I started just trying to narrow in on what we do know.

If we want a big quotient, we want the first mixed number to be big and the second to be small. To get that, it is not too hard to see that a, b, and f should be big numbers and c, d, and e should be small numbers. There isn’t any point to making any of them 4-6, because making any of them smaller/bigger would lead to a better result. So a, b, and f should be 7, 8, and 9 (not necessarily in that order) and c, d, and e should be 1, 2, and 3 (in some order).

But that’s a lot of possibilities to check out — 6 ways to pick a/b/f values and 6 for c/d/e, for 36 total, all with gory fraction division. Doable, but very tedious. So I tried to knock out some possibilities.

When I solved it, I first realized that making a big had to be at least as important than making b big, because a is the number of units and b is, at MOST, also the number of units (otherwise the number of halves, thirds, etc.). So we have nothing to lose by saying a must be bigger than b. Similarly, d should be smaller than e. This knocked the number of possibilities down to 9, and I solved them all to find the answer.

I thought the answer shown was likely correct, but it turns out several possibilities were larger. The very largest was 9 8/2 divided by 1 3/7, which is 9.1.

Upon seeing that, I managed to convince myself that I should have known the whole number in the first part had to be 9 and in the second part it had to be 1. But by light of day, I don’t think it’s so obvious, since making c small is a big deal, also. I’m not sure if last-night-me was onto something or not.

As the original answer implies, you just switch the mixed numbers to get the smallest quotient. This makes sense because maximizing the original quotient is like minimizing its reciprocal.

Very cool problem! It was captivating me when the homework I was supposed to be doing was evaluating the (dreadful, horrifying, at least the parts I looked at) LearnZillion online educational resources. No question which I preferred…

I saw you are listed in the NW Mathematics Conference program; I look forward to seeing you present there.

2. Robert Kaplinsky

Julie, what a fantastic write up of the problem! This is at the heart of what is wonderful about DOK 2 and 3 problems. I love the way you strategized your answer. Thank you so much for sharing it.

3. How cool to do a search on sixth grade number system lessons and come across this a month after I wrote it and had forgotten I had! Thanks for the positive feedback.

4. 8 9/1 divided by 7 6/2