AMC 10 · 2019 · #19
Grade 6 arithmeticProblem
Let S be the set of all positive integer divisors of 100,000. How many numbers are the product of two distinct elements of S?
Pick an answer.
AMC 10 2019 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.
Toolkit + CCSS Solution
Understand
Restated: Let $S$ be the set of all positive divisors of $100{,}000$. How many distinct numbers can be written as the product of two *distinct* elements of $S$?
Givens: $100{,}000 = 10^5 = 2^5 \cdot 5^5$; $S$ = set of all positive divisors of $100{,}000$; Each divisor has the form $2^a \cdot 5^b$ with $0 \le a \le 5$ and $0 \le b \le 5$; $|S| = 6 \cdot 6 = 36$; Choices: (A) $98$, (B) $100$, (C) $117$, (D) $119$, (E) $121$
Unknowns: The number of distinct values $d_1 \cdot d_2$ with $d_1, d_2 \in S$ and $d_1 \ne d_2$
Understand
Restated: Let $S$ be the set of all positive divisors of $100{,}000$. How many distinct numbers can be written as the product of two *distinct* elements of $S$?
Givens: $100{,}000 = 10^5 = 2^5 \cdot 5^5$; $S$ = set of all positive divisors of $100{,}000$; Each divisor has the form $2^a \cdot 5^b$ with $0 \le a \le 5$ and $0 \le b \le 5$; $|S| = 6 \cdot 6 = 36$; Choices: (A) $98$, (B) $100$, (C) $117$, (D) $119$, (E) $121$
Plan
Primary tool: #16 Change Focus / Count the Complement
Secondary: #7 Identify Subproblems, #9 Solve an Easier Related Problem
Tool #16 (Complement): list ALL candidate values (divisors of $10^{10}$, which is $11 \cdot 11 = 121$) and subtract the few that cannot be written as a product of two *distinct* divisors of $10^5$. Tool #7 (Subproblems): a product of two divisors has the form $2^a \cdot 5^b$ with $0 \le a, b \le 10$ — so the question splits into "which $(a, b)$ pairs are reachable?" Tool #9 (Easier Problem): the same kind of question on $100 = 2^2 \cdot 5^2$ first, then generalize.
Execute — Answer: C
6.NS.B.4 Step 1 - Set up the candidate universe.
- Each element of $S$ has the form $2^a \cdot 5^b$ with $0 \le a, b \le 5$.
- A product of two elements has exponents $a_1 + a_2 \in [0, 10]$ and $b_1 + b_2 \in [0, 10]$.
- So every product is a divisor of $2^{10} \cdot 5^{10}$, which has $(10+1)(10+1) = 121$ divisors.
- These $121$ values are the *candidate* products.
💡 Products live in a $11 \times 11$ grid of $(a, b)$ exponent sums — start with the whole grid.
6.EE.B.6 Step 2 - Now identify which candidates CANNOT be obtained from two *distinct* divisors.
- A value $2^A \cdot 5^B$ comes from distinct $(a_1, b_1) \ne (a_2, b_2)$ with $a_1 + a_2 = A$, $b_1 + b_2 = B$, and each of $a_i, b_i$ in $[0, 5]$.
- The constraint is: the chosen $(a_1, b_1)$ and $(a_2, b_2)$ must be different points in $[0, 5] \times [0, 5]$.
💡 Translate "distinct divisors" to "distinct $(a, b)$ pairs".
6.EE.B.6 Step 3 - Check $A = B = 0$, i.e., the candidate value $1$.
- We need $a_1 + a_2 = 0$ and $b_1 + b_2 = 0$ with $a_i, b_i \ge 0$ — forces $a_1 = a_2 = b_1 = b_2 = 0$, so both divisors are $1$.
- Not distinct.
- So $1$ is unreachable.
💡 Only way to multiply to $1$ is $1 \times 1$, but that's the same divisor twice.
6.EE.B.6 Step 4 - Check $A = 10, B = 10$, candidate $2^{10} \cdot 5^{10} = 10^{10}$.
- Need $a_1 + a_2 = 10$ with $a_i \le 5$, forcing $a_1 = a_2 = 5$; similarly $b_1 = b_2 = 5$.
- So both divisors are $2^5 \cdot 5^5 = 100{,}000$.
- Not distinct.
- Unreachable.
💡 Top corner of the grid has only one factorization — twin divisors.
6.EE.B.6 Step 5 - Check $A = 10, B = 0$, candidate $2^{10}$.
- Need $a_1 + a_2 = 10$ with $a_i \le 5$, so $a_1 = a_2 = 5$.
- And $b_1 + b_2 = 0$, so $b_1 = b_2 = 0$.
- Both divisors $= 2^5 \cdot 5^0 = 32$.
- Not distinct.
- Unreachable.
💡 Same forced-pair trap, just on the $2$-axis.
6.EE.B.6 Step 6 - Check $A = 0, B = 10$, candidate $5^{10}$.
- Same forced pair: both divisors $= 5^5 = 3125$.
- Not distinct.
- Unreachable.
💡 Mirror image of the previous case.
6.EE.B.6 Step 7 - Verify every OTHER candidate IS reachable.
- Take any $(A, B)$ in $[0, 10]^2$ not in the four bad corners.
- Strategy: pick $a_1 = \min(A, 5)$ and adjust slightly to break a tie.
- Concretely, for $(A, B) = (1, 0)$: $(a_1, b_1) = (0, 0), (a_2, b_2) = (1, 0)$ — distinct, both in range.
- For $(A, B) = (5, 5)$: $(0, 0)$ and $(5, 5)$ — distinct.
- For $(A, B) = (10, 5)$: forces $a_1 = a_2 = 5$, but $b_1 + b_2 = 5$ allows $b_1 = 0, b_2 = 5$ — distinct $(5, 0), (5, 5)$.
- The only candidates with NO distinct-pair choice are the four corners $(0,0), (10,0), (0,10), (10,10)$ — because each forces both coordinates to their forced single value.
💡 Force-the-pair fails only when BOTH coordinates have a unique factorization — and that's only at the four extreme corners.
4.OA.A.3 Step 8 - Subtract: reachable $= 121 - 4 = 117$.
- Answer: choice (C).
💡 Universe minus exceptions — done.
6.NS.B.4 Set up the candidate universe. Each element of $S$ has the form $2^a \cdot 5^b$ 6.EE.B.6 Now identify which candidates CANNOT be obtained from two *distinct* divisors. A 6.EE.B.6 Check $A = B = 0$, i.e., the candidate value $1$. We need $a_1 + a_2 = 0$ and $b 6.EE.B.6 Check $A = 10, B = 10$, candidate $2^{10} \cdot 5^{10} = 10^{10}$. Need $a_1 + a 6.EE.B.6 Check $A = 10, B = 0$, candidate $2^{10}$. Need $a_1 + a_2 = 10$ with $a_i \le 5 6.EE.B.6 Check $A = 0, B = 10$, candidate $5^{10}$. Same forced pair: both divisors $= 5^ 6.EE.B.6 Verify every OTHER candidate IS reachable. Take any $(A, B)$ in $[0, 10]^2$ not 4.OA.A.3 Subtract: reachable $= 121 - 4 = 117$. Answer: choice (C). Review
Reasonableness: $117$ is between choices $100$ and $119$, and exactly $121 - 4$, where $121$ is the count of divisors of $100{,}000^2$ and $4$ is the count of corners that need twin factorizations. Sanity check on a smaller case: divisors of $100 = 2^2 \cdot 5^2$ are $9$ in number; products of two distinct divisors live among divisors of $10^4$ ($25$ candidates); bad corners are the same four ($(0,0), (4,0), (0,4), (4,4)$); so the answer for $100$ would be $25 - 4 = 21$, which matches a direct enumeration.
Alternative: Tool #2 (Systematic List): for each candidate $(A, B)$ in $[0, 10]^2$, count factorizations into two distinct elements of $S$. For most $(A, B)$ there are multiple, and we just need at least one — so the four-corner exception calculation is the cleanest. Direct enumeration without the complement is much slower.
CCSS standards used (min grade 6)
4.OA.A.3Solve multi-step word problems using four operations with whole numbers (Final subtraction $121 - 4 = 117$.)6.NS.B.4Find greatest common factor and least common multiple of two numbers (Working with prime factorization $100{,}000 = 2^5 \cdot 5^5$ and counting divisors as $(5+1)(5+1)$.)6.EE.B.6Use variables to represent numbers and write expressions to solve problems (Writing divisors as $2^a \cdot 5^b$ and reducing the question to which $(A, B)$ exponent sums are reachable from distinct $(a_i, b_i)$ pairs.)
⭐ This AMC 10 problem only needs Grade 6 prime-factorization reasoning you already know — every divisor of $10^5$ is $2^a \cdot 5^b$, so products live among the $121$ divisors of $10^{10}$, and only $4$ "corner" values fail the distinct-pair rule. $121 - 4 = 117$. The answer is $\textbf{(C)}$.
⭐ This AMC 10 problem only needs Grade 6 prime-factorization reasoning you already know — every divisor of $10^5$ is $2^a \cdot 5^b$, so products live among the $121$ divisors of $10^{10}$, and only $4$ "corner" values fail the distinct-pair rule. $121 - 4 = 117$. The answer is $\textbf{(C)}$.
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