AMC 10 · 2008 · #3

Easy mode Grade 4

Problem

For a counting number $n$ , let $\langle n\rangle$ mean this: add up every number that divides $n$ evenly, but leave out $n$ itself. For example, $\langle 4\rangle=1+2=3$ , and $\langle 12 \rangle =1+2+3+4+6=16$ . Work from the inside out. What is $\langle\langle\langle 6\rangle\rangle\rangle$ ?

Pick an answer.

(A)

$\ 6$

(B)

$\ 12$

(C)

$\ 24$

(D)

$\ 32$

(E)

$\ 36$

AMC 10 2008 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

Try it yourself first — the explanation is most useful after you’ve attempted it.

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Toolkit + CCSS Solution

Understand

Restated: The symbol \langle n\rangle means: add up every positive divisor of n except n itself. Apply this operation three times in a row, starting from 6, and find the final result \langle\langle\langle 6\rangle\rangle\rangle.

Givens: \langle n\rangle is the sum of all positive divisors of n except n itself.; Example: \langle 4\rangle = 1 + 2 = 3.; Example: \langle 12\rangle = 1 + 2 + 3 + 4 + 6 = 16.; The starting number is 6.

Unknowns: The value of \langle\langle\langle 6\rangle\rangle\rangle, the operation applied three times.

Understand

Plan

Primary tool: #7 Identify Subproblems

Secondary: #2 Make a Systematic List, #5 Look for a Pattern

The triple brackets are three separate \langle\ \rangle operations stacked on each other. Evaluate the innermost one first, then feed its result into the next. Each \langle\ \rangle is itself a small task: list the proper divisors, then add them.

Execute — Answer: A

#2 Make a Systematic List 4.OA.B.4 Step 1

List the divisors of 6

Start inside the brackets with \langle 6\rangle.
Find every positive integer that divides 6 evenly: 1, 2, 3, and 6.
The definition of \langle\ \rangle throws out the number itself, so drop the 6 and keep the proper divisors 1, 2, 3.

$$6 = 1\times 6 = 2\times 3 \Rightarrow \text{divisors } 1,2,3,6;\ \text{proper divisors } 1,2,3$$

💡 Every divisor pairs with a partner that multiplies back to 6, so checking up to 3 already finds them all.

#7 Identify Subproblems 1.OA.C.6 Step 2

Add them to get the value of 6

Add the proper divisors to finish the innermost operation: 1 + 2 + 3 = 6, so \langle 6\rangle = 6.
The number 6 returns itself, which makes it a perfect number.

$$\langle 6\rangle = 1+2+3 = 6$$

💡 A perfect number equals the sum of its own proper divisors, so \langle 6\rangle lands right back on 6.

#5 Look for a Pattern 4.OA.C.5 Step 3

Apply the operation again

Now the expression is \langle\langle 6\rangle\rangle = \langle\,\langle 6\rangle\,\rangle.
Because \langle 6\rangle already equals 6, this middle bracket is just \langle 6\rangle once more, which is 6 again.
The value does not change.

$$\langle\langle 6\rangle\rangle = \langle 6\rangle = 6$$

💡 Once a number maps to itself, repeating the rule keeps handing back the same number.

#5 Look for a Pattern 4.OA.C.5 Step 4

Apply the operation a third time

The outer bracket does the same thing one more time: \langle\langle\langle 6\rangle\rangle\rangle = \langle 6\rangle = 6.
Applying the operation any number of times to a perfect number leaves it unchanged, so the final result is 6, which is answer (A).

$$\langle\langle\langle 6\rangle\rangle\rangle = \langle 6\rangle = 6$$

💡 6 is a fixed point of \langle\ \rangle, so nesting the brackets to any depth still gives 6.

[1] #2 4.OA.B.4 Start inside the brackets with \langle 6\rangle. Find every positive integer tha

[2] #7 1.OA.C.6 Add the proper divisors to finish the innermost operation: 1 + 2 + 3 = 6, so \la

[3] #5 4.OA.C.5 Now the expression is \langle\langle 6\rangle\rangle = \langle\,\langle 6\rangle

[4] #5 4.OA.C.5 The outer bracket does the same thing one more time: \langle\langle\langle 6\ran

Review

Reasonableness: The answer 6 matches the starting number, which fits the key fact that 6 is a perfect number (1+2+3=6). Since each \langle\ \rangle maps 6 to 6, the triple nesting must also give 6. Choice (A) is 6, while the larger choices 12, 24, 32, and 36 would require the value to grow, which cannot happen at a fixed point.

Alternative: Instead of spotting the perfect-number shortcut, compute all three brackets mechanically: \langle 6\rangle = 1+2+3 = 6, then \langle 6\rangle = 6 again, then \langle 6\rangle = 6 a third time. The brute-force route lands on the same 6.

CCSS standards used (min grade 4)

4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite (Listing every positive divisor of 6 to identify its proper divisors)
1.OA.C.6 Add and subtract within 20 using strategies (Summing the proper divisors 1 + 2 + 3 to evaluate \langle 6\rangle)
4.OA.C.5 Generate a number or shape pattern following a given rule (Recognizing that \langle 6\rangle = 6 repeats, so each further bracket leaves the value unchanged)

⭐ 6's proper divisors add back up to 6, so no matter how many times you take the divisor-sum, you always land on 6.

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: A

List the divisors of 6

Add them to get the value of 6

Apply the operation again

Apply the operation a third time

Review

CCSS standards used (min grade 4)

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