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The following is an elementary-level Math Kangaroo multiple choice question:

What is the maximum value of the sum of the digits of the sum of the digits of a three-digit number?

A. 9
B. 10
C. 11
D. 12
E. 18

I have the answer key, so I can work backward to sort of get how to solve this problem. However, as I was trying to write a solution for my students, I realized I could NOT help them eliminate the wrong answer choices. The following is what I have got so far:

This is a tricky problem. You might have no glue the first time reading the question, but you can still capture some keywords such as “maximum” and “three-digit number”. These keywords can give you some information to start solving the problem.

When you saw the word “maximum”, “three-digit number”, you could started trying out with the biggest three digit-number, which is 999. Now you can plug in the number into the question, then you will find out the question is asking “What is the maximum value of the sum of the digits of the sum of the digits of 999?”

Now, the question is getting clearer. First, you need to solve the sum of the three-digits number 999, which is 9 + 9 + 9 = 18 + 9 = 27. The question turns into “What is the maximum value of the sum of the digits of 27”. Then you need to solve the sum of 27, which is 2+7 = 9.

Take a look at the answer choices; there is a 9. This is a signal that you are on the right track, but before you circle the answer, don’t forget the power of double check! You need to ask yourself, is nine the maximum value you can get from combining two single digits?

Now I am stuck with how to help the student eliminate other answer choices.

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    $\begingroup$ Are the highlighted passages something you would suggest to tell a student? They contain some mistakes themselves, like What is the maximum value (...) of 999? (which makes no sense), you need to solve the sum (you don't solve a sum, you compute it, so that's confusing), This is a signal that you are on the right track (why? it looks like the wrong track, actually). $\endgroup$ Commented Jan 5, 2023 at 2:23
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    $\begingroup$ @SueVanHattum: 469 => 19 => 10. It's not 9!!! $\endgroup$
    – Dominique
    Commented Jan 5, 2023 at 9:13
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    $\begingroup$ @Dominique The way the post is written, it's not clear what the poster thinks the right answer is. By making some wrong assumptions, they arrive at an answer of 9. But then they suggest double checking, indicating that perhaps they know 9 isn't right. On the other hand, they say they want to "eliminate other answer choices", hinting that maybe they think 9 is right. By the way, spending time trying to eliminate wrong answer choices doesn't seem like an effective strategy for this problem. $\endgroup$ Commented Jan 5, 2023 at 13:50
  • $\begingroup$ This question doesn't even make sense. I can make the digit sum as big as I like, just change the base :) $\endgroup$ Commented Jan 7, 2023 at 14:34

3 Answers 3

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Since digits are less than or equal to 9, then you know the sum of the digits of a 3 digit number is less than or equal to 27. So the question becomes "what number less than or equal to 27 has the largest sum of digits?" Since 27 has a digit-sum of 9, you look for a number with digit-sum 10 or more. 19 has a digit-sum of 10. You know that a digit-sum of 11 or more is impossible since the first digit is no more than 2 (and 27 is the highest). Thus the answer is B. 10. (Note: to check that this max actually occurs, inspection finds 667 gives digit-sum 19 which has digit-sum 10.)

Edit (from the comments): To get students there I would ask them to (i) find the biggest digit-sum from a 2-digit number less than various 2-digit numbers (such as 72, 58, or 40), (ii) find the biggest digit-sum from a 3-digit number, (iii) solve the original question.

Edit (about English language difficulties mentioned in the comments): At an older level the ideas can be bridged by using function notation. Defining $s(n)=\text{sum of the digits of $n$}$, e.g. $s(87)=8+7=15$ and so the question is asking to maximize $s(s(n))$.

At the elementary level, using an arrow notation for functions instead usually gets the point across. So explain that an arrow gives sum of the digits, e.g. $87 \Rightarrow 8+7 = 15$ and then asking if you start with a three-digit number what's the largest number you can get in the last blank of: $$\underline{\ \ \ \ } \Rightarrow \underline{\ \ \ \ \ } =\underline{\ \ \ \ \ } \Rightarrow \underline{\ \ \ \ \ }=\underline{\ \ \ \ \ } .$$ It would also be good to do an example like $$\underline{\ 147\ } \Rightarrow \underline{\ 1+4+7\ } =\underline{\ \ 12\ \ } \Rightarrow \underline{\ 1+2\ }=\underline{\ \ 3\ \ } .$$

When I've done similar things with elementary students, they usually find the correct maximum pretty quickly through trial and error. So then you need to push them to explain why they are sure it's the largest (without checking every three-digit number).

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    $\begingroup$ 667, along with many others, including 991. $\endgroup$
    – Sue VanHattum
    Commented Jan 5, 2023 at 5:24
  • $\begingroup$ To find a number < 27 with higher digit sum, you need to lower the first digit a step and see if there are any possible second digits that are two higher than 7. To find a three digit number with digit sum 19, you just find any three numbers a,b,c < 10 that sum to 19. For a = zero the max is 18, for any a 1 to 9 you just subtract it from 17 and the others sum to that. $\endgroup$
    – JollyJoker
    Commented Jan 5, 2023 at 12:32
  • $\begingroup$ @SueVanHattum I noticed that if you have one solution then adding 198 will give another solution. $\endgroup$ Commented Jan 5, 2023 at 13:32
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    $\begingroup$ @StevenGubkin Either adding 99 (increasing 100s digit by 1 while decreasing 1s digit by 1) or adding 90 (increasing 100s digit by 1, while decreasing 10s digit by 1) works. $\endgroup$ Commented Jan 5, 2023 at 14:16
  • $\begingroup$ @willorrick nice observation! $\endgroup$ Commented Jan 5, 2023 at 14:41
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You don't need to know just the largest digit sum of three digit numbers, but all of them. 999 has the largest digit sum 27, 100 has the smallest digit sum 1, and everything in between is possible. So the first digit sum is anything from 1 to 27.

The digit sum of 27 is 9. But that is not the largest digit sum. We can pick a smaller number, specifically 19, with a larger digit sum. That's because by making the first digit 1 smaller, we can make the second digit two larger. This gives a digit sum of 10.

As a bonus, we get the first number with a digit sum of 19 if we make the tens and single digits as large as possible and the hundreds digit as small as possible, which is the number 199.

As a bonus, which is the smallest number where the sum of digits of the sum of digits is larger, that is 11? The smallest number with a digit sum of 11 is a digit 9 and a digit 2, that's 29. And the smallest number with a digit sum of 29 has three nine's and a 2, that is 2,999.

And a third bonus question: Which is the smallest number where E is the right answer? E means: Sum of digits of the sum of digits is 18. That's two nine's, so the sum of digits is 99. That's an enormous number: We need eleven nines, so the smallest number is 99,999,999,999.

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    $\begingroup$ The question is about how to help students get there, not what the right answer is. Just telling them the right answer is unlikely to be very effective in helping them learn. $\endgroup$
    – Tommi
    Commented Jan 5, 2023 at 13:31
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    $\begingroup$ Thank you for sharing this; it is really helpful. What I was trying to do is to help the students understand what the question is asking first; most of our students are native Chinese speakers, so even the word order is confusing, but I think I could find a way to combine the answer you provided and revise what I've gotten so far to provide a comprehensive solution to my students. $\endgroup$
    – Tessa
    Commented Jan 5, 2023 at 15:23
  • $\begingroup$ @Tommi, "The question is about how to help students get there" It seems to me that gnasher729 explained in detail how to get the answer. If this explanation was presented to the students, then I believe this can be effective in helping them learn. $\endgroup$
    – JRN
    Commented Jan 7, 2023 at 1:31
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I understand your question to target the "average student", not the top ones (who will most probably find the solution more or less by intuition). So, the focus lies on explaining a failsafe way that can be easily understood.

Have them do it in tiny, explicit steps.

maximum value of the sum of the digits of the sum of the digits of a three-digit number

Analyzing a phrase with so many "of" words starts at the end of the phrase.

  • Start with "three-digit number" and ask about the possible numbers matching that expression, leading to "anything from 100 to 999".

  • Go to "the sum of the digits of a three-digit number". The basic, brute-force approach is to calculate all the hundreds of digit-sums, but as any combination of the digits 1-9, 0-9, 0-9 is possible, we can take a shortcut to convince ourselves that anything from 1+0+0 to 9+9+9 is possible, so the sum can be anything between 1 and 27, inclusive.

  • For "the sum of the digits of the sum of the digits of a three-digit number", being "the sum of the digits of a number between 1 and 27", taking a shortcut is tricky, as there are different cases to be taken care of: one-digit numbers, two-digit numbers beginning with a 1, and two-digit numbers beginning with a 2. So, writing down the 27 numbers and calculating their sum of digits is doable and will lead to a list of the possible sum-of-digits values, ranging from 1 to 10.

  • Clearly, the "maximum value of ..." then is 10.

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