AMC 10 · 2013 · #20

Grade 6 number-theory

Archetype Small-Domain Constrained Search └ Sub-archetype Factor / Divisibility Constraint

prime-factorizationfactorialprime-numbers extremal-constructionidentify-subproblems ↑ Prerequisites: prime-factorizationfactorial

📏 Medium solution 💡 3 insights

Problem

The number $2013$ is expressed in the form

$2013 = \frac {a_1!a_2!...a_m!}{b_1!b_2!...b_n!}$ ,

where $a_1 \ge a_2 \ge \cdots \ge a_m$ and $b_1 \ge b_2 \ge \cdots \ge b_n$ are positive integers and $a_1 + b_1$ is as small as possible. What is $|a_1 - b_1|$ ?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 10 2013 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: Write 2013 as a fraction whose top and bottom are each products of factorials, 2013 = (a_1! a_2! ... a_m!)/(b_1! b_2! ... b_n!), where the a's are non-increasing positive integers and the b's are non-increasing positive integers. Among all such ways, pick one where a_1 + b_1 is as small as possible. Report |a_1 - b_1|.

Givens: 2013 must equal a product of factorials on top divided by a product of factorials on the bottom; a_1 >= a_2 >= ... >= a_m are positive integers (a_1 is the largest number on top); b_1 >= b_2 >= ... >= b_n are positive integers (b_1 is the largest number on the bottom); Among all valid expressions, a_1 + b_1 is minimized

Unknowns: The value of |a_1 - b_1| for the a_1 + b_1 minimizing expression

Understand

Plan

Primary tool: #14 Extreme Principle

Secondary: #7 Identify Subproblems, #6 Guess and Check

The phrase 'as small as possible' points straight at the Extreme Principle: chase the smallest a_1 and the smallest b_1 that can still work. The lever is prime factors. A factorial n! contains a prime p only if n reaches p, so the largest prime inside 2013 forces a_1, and the largest unwanted prime dragged along forces b_1. Break the problem into 'how small can a_1 be' and 'how small can b_1 be', then build one explicit expression to prove those minimums are reachable.

Execute — Answer: B

#7 Identify Subproblems 4.OA.B.4 Step 1

Factor 2013 into primes

Break 2013 into prime factors.
2013 is divisible by 3, giving 2013 = 3 x 671, and 671 = 11 x 61.
Both 11 and 61 are prime, so 2013 = 3 x 11 x 61.
The largest prime factor is 61.

$$2013 = 3 \times 11 \times 61$$

💡 Splitting into primes shows exactly which building blocks the factorials must supply.

#14 Extreme Principle 6.NS.B.4 Step 2

The prime 61 forces a_1 = 61

Since 61 divides 2013, the top product must be divisible by 61.
A factorial n!
is divisible by the prime 61 only when n is at least 61.
So at least one number on top is 61 or more, meaning the largest, a_1, is at least 61.
To keep a_1 as small as possible, take a_1 = 61.

$$61 \mid 2013 \;\Rightarrow\; a_1 \ge 61, \qquad \min a_1 = 61$$

💡 A prime can only enter a factorial once the count climbs up to that prime.

#14 Extreme Principle 6.NS.B.4 Step 3

61! drags in 59, forcing b_1 = 59

With a_1 = 61 the top contains 61!, and 61!
includes the prime 59 (the largest prime below 61, since 60 is not prime).
But 59 does not divide 2013, so that unwanted 59 must be cancelled by the bottom.
A factorial holds the prime 59 only when its number is at least 59, so the largest bottom number b_1 must be at least 59.
Taking b_1 = 59 makes a_1 + b_1 = 61 + 59 = 120, and no choice can beat that since a_1 >= 61 already and b_1 >= 59 whenever a_1 = 61.

$$61! \text{ contains } 59,\; 59 \nmid 2013 \;\Rightarrow\; b_1 \ge 59, \qquad a_1 + b_1 = 120$$

💡 Whatever unwanted prime the top factorial pulls in must be matched by a factorial on the bottom to erase it.

#6 Guess and Check 5.OA.A.1 Step 4

Build an expression that hits 120

Check that a_1 = 61 and b_1 = 59 are actually achievable.
Start from 61!/59!
= 60 x 61 = 3660 = 2^2 x 3 x 5 x 61.
Multiplying by 11!/10!
= 11 and by 3!/5!
= 1/20 gives 3660 x 11 / 20 = 2013.
So 2013 = (61!
11!
3!)/(59!
10!
5!), with the tops 61 >= 11 >= 3 and bottoms 59 >= 10 >= 5 all in order.
This is a valid expression reaching the minimum a_1 + b_1 = 120.

$$\frac{61!\,11!\,3!}{59!\,10!\,5!} = (60\cdot 61)\cdot 11 \cdot \frac{1}{20} = 3660\cdot\frac{11}{20} = 2013$$

💡 Showing one working expression proves the smallest a_1 and b_1 are really possible, not just wished for.

#7 Identify Subproblems 4.NBT.B.4 Step 5

Take the difference

With a_1 = 61 and b_1 = 59 fixed by the minimum, the requested quantity is |a_1 - b_1| = |61 - 59| = 2.
So the answer is (B).

$$|a_1 - b_1| = |61 - 59| = 2$$

💡 The two forced anchors sit just two apart because 59 is the prime right below 61.

[1] #7 4.OA.B.4 Break 2013 into prime factors. 2013 is divisible by 3, giving 2013 = 3 x 671, an

[2] #14 6.NS.B.4 Since 61 divides 2013, the top product must be divisible by 61. A factorial n! i

[3] #14 6.NS.B.4 With a_1 = 61 the top contains 61!, and 61! includes the prime 59 (the largest p

[4] #6 5.OA.A.1 Check that a_1 = 61 and b_1 = 59 are actually achievable. Start from 61!/59! = 6

[5] #7 4.NBT.B.4 With a_1 = 61 and b_1 = 59 fixed by the minimum, the requested quantity is |a_1

Review

Reasonableness: The two anchors come from opposite ends: 61 is the largest prime inside 2013 (so it must appear on top), and 59 is the largest prime that 61! unavoidably drags along but 2013 does not want (so it must appear on the bottom). Both bounds a_1 >= 61 and b_1 >= 59 are tight, and the explicit expression (61! 11! 3!)/(59! 10! 5!) = 2013 confirms both are reached, so |61 - 59| = 2 is solid.

Alternative: Instead of building the witness expression, argue by pure necessity: any valid form needs 61 on top (else no factor of 61) giving a_1 >= 61, and once 61! is present the prime 59 it carries has nowhere to go unless some bottom factorial reaches 59, giving b_1 >= 59. Since a_1 + b_1 >= 120 always and equality is attainable, the minimizing pair is (61, 59) and the difference is 2.

CCSS standards used (min grade 6)

4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite (Breaking 2013 into 3 x 11 x 61 and spotting 61 as its largest prime factor.)
6.NS.B.4 Find greatest common factor and least common multiple of two numbers (Reasoning about which primes a factorial contains to force a_1 = 61 and b_1 = 59.)
5.OA.A.1 Use parentheses, brackets, or braces in numerical expressions and evaluate (Evaluating (61! 11! 3!)/(59! 10! 5!) to confirm it equals 2013.)
4.NBT.B.4 Fluently add and subtract multi-digit whole numbers (Computing a_1 + b_1 = 120 and the final |61 - 59| = 2.)

⭐ The biggest prime inside a number decides the top factorial, and the biggest unwanted prime it drags along decides the bottom one.

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: B

Factor 2013 into primes

The prime 61 forces a_1 = 61

61! drags in 59, forcing b_1 = 59

Build an expression that hits 120

Take the difference

Review

CCSS standards used (min grade 6)

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