Please don’t solve this problem for me

I’ve been reading Tracy Zager’s Becoming the Math Teacher You Wish You’d Had. (Full disclosure: Tracy is my editor at Stenhouse, which published Which One Doesn’t Belong?)

In Chapter 3: Mathematicians Take Risks, Tracy invites us to follow her lead and take a risk in a manner similar to a student she documents in a classroom vignette. She suggests that we play with 10. I liked the idea, but was feeling pretty tapped out on risks related to 10, having explored that territory in great detail and complexity over the last ten years. So I read on with Tracy’s challenge in the back of my mind.

In a seemingly unrelated episode, Tabitha (9 years old) was doing some homework this week and needed to know the product 7 times 8. Spouse was offering a hint: It starts with a five. I asked Tabitha if she knew what 10 sevens is, which led to her expressing 7 x 8 as 70 – 14.

The next day I was thinking about Spouse’s hint. It starts with a 5. What math might there be in that relationship? Could I develop a way of knowing what a product starts with?

With Tracy’s voice in my head, I took a risk. I now take a bigger risk by sharing my ideas with you.


If you’re multiplying 7 by 8, the product starts with a 5. This is two less than the multiplier, 7. The same is true for 8 times 8 (starts with 6) and 9 times 8 (starts with 7).

11 times 8 starts with an 8, though. If you look more deeply, you’ll notice that the difference between the tens digit (what the number starts with) and the multiplier increases by 1 for every group of 5 in the multilplier.

Try 22 times 8. 22 has four groups of 5, so 22 times 8 starts with 17 (which is 22-4, minus one more because 1 times 8 starts with a zero).

The opposite rule applies to twelves. Every five twelves, you need to ADD an extra unit to what the number starts with. 13 times 12 starts with 15 because there are two groups of five in 13.

There are exceptions. 25 times 8 has me stumped. It should start with 19, but does start with 20.


This is much more complicated. You can get a good approximation by counting groups of 10 in the multiplier and adding three to the difference between starting digits and multiplier for each group of 10. But you’ll get a lot of exceptions.

71 works. Seven 10s, times three gives you 21, plus one is 22. 71-22 is 49, so we expect 71 x 7 to start with 49, and it does: 497.

But 78 fails. My rule predicts that 78 x 7 starts with 56, but it doesn’t. 78 x 7 = 546.

The failures seem to be predictable, though. So I’ll press onwards.

I haven’t yet decided whether 7 is different and difficult because of its 7-ness, or because of the fact that it’s 3 away from 10. The three for every ten bit makes me think it’s probably the threeness of seven that matters rather than the sevenness. But I don’t know that yet.

Please don’t answer these questions for me. They are mine. Explore them if you find them compelling. Don’t spoil my fun.


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