AMC 10 · 2015 · #6
Easy mode Grade 5Tillie's multiplication table has one row for each of the labels 0,1,2,3,4,5,6,7,8,9,10,11,12, and one column for each of the same labels. That is 13 labels, so the table has 13 rows and 13 columns, counting the row for 0 and the column for 0. Each cell inside the table holds its row label times its column label. What fraction of the numbers inside the table are odd? Write your answer as a decimal rounded to the nearest hundredth.
Pick an answer.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: A multiplication table has its rows and its columns labeled with the whole numbers from $0$ up to $12$. The body of the table holds the product of the row label and the column label in every cell. Find what share of those cells hold an odd number, written as a decimal rounded to the nearest hundredth.
Givens: The facts run from $0 \times 0$ to $12 \times 12$, so the row labels are $0, 1, 2, \ldots, 12$ and the column labels are the same list.; The body of the table is every cell where a row meets a column, and each such cell holds the product of the two labels.; The labels themselves sit outside the body, so they are not counted as entries.
Unknowns: The number of cells in the body of the table whose entry is odd.; That count divided by the total number of cells, expressed as a decimal to the nearest hundredth.
Understand
Restated: A multiplication table has its rows and its columns labeled with the whole numbers from $0$ up to $12$. The body of the table holds the product of the row label and the column label in every cell. Find what share of those cells hold an odd number, written as a decimal rounded to the nearest hundredth.
Givens: The facts run from $0 \times 0$ to $12 \times 12$, so the row labels are $0, 1, 2, \ldots, 12$ and the column labels are the same list.; The body of the table is every cell where a row meets a column, and each such cell holds the product of the two labels.; The labels themselves sit outside the body, so they are not counted as entries.
Plan
Primary tool: #2 Make a Systematic List
Secondary: #1 Draw a Diagram, #5 Look for a Pattern, #9 Solve an Easier Related Problem, #16 Change Focus / Count the Complement, #3 Eliminate Possibilities
Nobody wants to compute $169$ products. The question only asks whether each entry is odd, and whether a product is odd depends on nothing but the parity of the two labels. So the plan is: first pin down exactly how many cells the body has, then prove the parity rule for products in both directions, then count the odd labels and multiply. Because the whole result rests on two claims that are easy to state and easy to get wrong — that the body has $13$ rows and not $12$, and that a product is odd exactly when both factors are odd — each gets its own proof rather than being assumed. A complement count and a size bound on the answer choices then check the result along two routes that do not reuse the same arithmetic.
Execute — Answer: A
5.NBT.B.5 Step 1 Count the cells in the body
- Sketch the table: a strip of row labels down the left, a strip of column labels across the top, and the body filling the rectangle between them.
- The facts run from $0 \times 0$ to $12 \times 12$, so every label from $0$ to $12$ appears once on each strip.
- That list is $0, 1, 2, \ldots, 12$, which has $12 - 0 + 1 = 13$ numbers.
- Every row meets every column in exactly one cell, so the body is a $13$ by $13$ rectangle of cells, giving $13 \times 13 = 169$ entries.
- The $0$ row and the $0$ column are real rows and columns of the body: they hold the products $0 \times 0$ through $0 \times 12$, which are entries like any other.
- Dropping them would leave a $12$ by $12$ body of $144$ cells, and that is a different table from the one described.
💡 A list that starts at $0$ and ends at $12$ holds $13$ numbers, so the body is one row and one column bigger than the name "$12$ times table" makes it sound.
3.OA.B.5 Step 2 Prove when a product is odd
- Claim: for whole numbers $a$ and $b$, the product $a \times b$ is odd exactly when $a$ is odd and $b$ is odd.
- Both directions need checking, because the count uses both.
- First, suppose at least one factor is even, say $a = 2m$.
- Then $a \times b = 2m \times b = 2 \times (m \times b)$, which is two times a whole number, so it is even.
- The same argument works if $b$ is even.
- So an even factor forces an even product, which is what rules cells out.
- Second, suppose both factors are odd, say $a = 2m + 1$ and $b = 2n + 1$.
- Then $a \times b = 4mn + 2m + 2n + 1 = 2 \times (2mn + m + n) + 1$, which is one more than an even number, so it is odd.
- That is what rules cells in.
- Together the two directions say the odd cells are exactly the cells where both labels are odd — no more, no fewer.
💡 A single factor of $2$ anywhere in a product survives to the end, and if no factor carries one, nothing can create it.
2.OA.C.3 Step 3 List the odd labels
- Now scan the label list $0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12$ and pull out the odd ones: $1, 3, 5, 7, 9, 11$.
- That is $6$ odd labels.
- The other $7$ labels, $0, 2, 4, 6, 8, 10, 12$, are even — note that $0$ belongs on the even side, because $0 = 2 \times 0$, so the $0$ row and the $0$ column contribute no odd entries at all even though they do add to the total.
- The two strips carry the same list, so there are $6$ odd rows and $6$ odd columns.
💡 In a run of consecutive whole numbers starting and ending on an even number, the evens outnumber the odds by exactly one.
3.OA.A.1 Step 4 Count the odd cells
- By the rule just proved, a cell is odd exactly when its row label is odd and its column label is odd.
- So the odd cells form their own small rectangle inside the body: $6$ odd rows crossed with $6$ odd columns.
- Each odd row contributes one odd cell for each odd column, so each of the $6$ odd rows holds $6$ odd entries, and every even row holds none.
- That gives $6 \times 6 = 36$ odd entries in total.
- Test the method on a table small enough to check by hand: labels $0, 1, 2$ give the body $0, 0, 0, 0, 1, 2, 0, 2, 4$, which has exactly one odd entry, and the rule predicts $1 \times 1 = 1$ because only the label $1$ is odd.
- The method holds up.
💡 Odd rows crossed with odd columns is itself a multiplication table, so the odd cells are a $6$ by $6$ block hiding inside the $13$ by $13$ one.
4.NBT.B.4 Step 5 Check by counting the evens
- Before dividing, confirm the count of $36$ by a route that shares no arithmetic with it: count the even entries instead and subtract.
- An even entry appears wherever at least one label is even.
- Split that into two non-overlapping cases.
- Case one: the row label is even.
- There are $7$ such rows, and every one of their $13$ cells is even, giving $7 \times 13 = 91$ entries.
- Case two: the row label is odd but the column label is even.
- There are $6$ such rows and $7$ such columns, giving $6 \times 7 = 42$ entries.
- The cases cannot overlap, because a row label is either even or odd, and together they cover every even entry.
- So there are $91 + 42 = 133$ even entries, leaving $169 - 133 = 36$ odd ones.
- The two counts agree.
💡 Counting what you do not want and subtracting is a genuinely different computation, so agreement between the two is real evidence and not just the same slip repeated.
5.NBT.A.4 Step 6 Round the fraction without dividing
- The share of odd entries is $\frac{36}{169}$.
- Rounding to the nearest hundredth means deciding which hundredth it sits closest to, and that only needs to know which two hundredths it lies between — the halfway marks are $0.205 = \frac{41}{200}$ and $0.215 = \frac{43}{200}$.
- Compare by cross-multiplying, which keeps everything in whole numbers.
- Against the lower mark: $36 \times 200 = 7200$ and $41 \times 169 = 6929$, and $7200 > 6929$, so $\frac{36}{169} > 0.205$.
- Against the upper mark: $36 \times 200 = 7200$ and $43 \times 169 = 7267$, and $7200 < 7267$, so $\frac{36}{169} < 0.215$.
- The value is trapped strictly between the two halfway marks surrounding $0.21$, so it rounds to $0.21$ — no long division needed, and no digits to misread.
💡 To round a fraction you never need its digits, only which side of the two nearest halfway marks it falls on.
4.NF.A.2 Step 7 Confirm against the choices
- One last check that does not reuse the count.
- The odd labels are $6$ out of $13$, and $\frac{6}{13} < \frac{1}{2}$ because $6 \times 2 = 12$ is less than $13$.
- An odd entry needs an odd row and an odd column, so the share of odd entries is $\frac{6}{13} \times \frac{6}{13}$, which is less than $\frac{1}{2} \times \frac{1}{2} = \frac{1}{4} = 0.25$.
- So the answer must be strictly below $0.25$, and among the five choices only $0.21$ qualifies — $0.25$ itself is exactly what you would get from a $12$ by $12$ body with the zeros thrown away, which is the wrong table.
- This matches the computed value of $\frac{36}{169} \approx 0.21$, so the answer is $\textbf{(A)}\; 0.21$.
💡 Adding the even label $0$ stretches the denominator without adding a single odd entry, so it can only drag the share down.
5.NBT.B.5 Sketch the table: a strip of row labels down the left, a strip of column labels 3.OA.B.5 Claim: for whole numbers $a$ and $b$, the product $a \times b$ is odd exactly wh 2.OA.C.3 Now scan the label list $0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12$ and pull out 3.OA.A.1 By the rule just proved, a cell is odd exactly when its row label is odd and its 4.NBT.B.4 Before dividing, confirm the count of $36$ by a route that shares no arithmetic 5.NBT.A.4 The share of odd entries is $\frac{36}{169}$. Rounding to the nearest hundredth 4.NF.A.2 One last check that does not reuse the count. The odd labels are $6$ out of $13$ Review
Reasonableness: The value $\frac{36}{169} \approx 0.213$ sits between $0$ and $1$, as any share must, and it is a little under $\frac{1}{4}$ — exactly where it should be, since slightly fewer than half the labels are odd and an odd entry needs two of them at once. A direct spot check backs it up: the row labeled $7$ holds $0, 7, 14, 21, 28, 35, 42, 49, 56, 63, 70, 77, 84$, of which $7, 21, 35, 49, 63, 77$ are odd — six odd entries, matching the prediction of one per odd column, while the row labeled $8$ holds only even entries. The two answer traps are visible too: forgetting the zero labels gives $\frac{36}{144} = 0.25$, choice (B), and counting the odd labels as half of all labels gives $0.25$ again, so any route that lands on $0.25$ has quietly dropped the zero row and column.
Alternative: Replace every entry with its parity and read the table as a pattern instead of a count. Writing $1$ for odd and $0$ for even, the parity table is the outer product of the label parity strip $0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0$ with itself, so the $1$s appear exactly where a $1$ meets a $1$ — a checkerboard of isolated $1$s occupying every other row and every other column. Counting $1$s in such a grid is just $6 \times 6 = 36$ again, but it arrives visually rather than arithmetically. A third route is probabilistic: pick a cell at random, and the chance its row label is odd is $\frac{6}{13}$, independently the same for its column, so the chance the entry is odd is $\frac{6}{13} \times \frac{6}{13} = \frac{36}{169}$. All three routes land on the same fraction.
CCSS standards used (min grade 5)
2.OA.C.3Determine whether a group of objects has an odd or even number (Sorting the labels $0$ through $12$ into $6$ odd and $7$ even, including the recognition that $0$ is even.)3.OA.A.1Interpret products of whole numbers as total number of objects in groups (Reading the odd cells as $6$ odd rows each holding $6$ odd entries, so the count is $6 \times 6 = 36$.)3.OA.B.5Apply properties of operations as strategies to multiply and divide (Factoring $ab = 2(mb)$ when $a = 2m$, and expanding $(2m+1)(2n+1) = 2(2mn+m+n)+1$, to prove a product is odd exactly when both factors are odd.)4.NBT.B.4Fluently add and subtract multi-digit whole numbers (The complement check: $91 + 42 = 133$ even entries and $169 - 133 = 36$ odd ones.)4.NF.A.2Compare two fractions with different numerators and different denominators (Showing $\frac{6}{13} < \frac{1}{2}$, which forces the share of odd entries below $\frac{1}{4}$ and eliminates every choice except $0.21$.)5.NBT.A.4Round decimals to any place (Deciding that $\frac{36}{169}$ rounds to $0.21$ by trapping it between the halfway marks $\frac{41}{200}$ and $\frac{43}{200}$.)5.NBT.B.5Fluently multiply multi-digit whole numbers (Finding the size of the body of the table as $13 \times 13 = 169$ cells.)
⭐ A product is odd only when every factor is odd, so counting odd entries in a times table is just counting odd row labels times odd column labels — and the row and column for $0$ still count as part of the table.
⭐ A product is odd only when every factor is odd, so counting odd entries in a times table is just counting odd row labels times odd column labels — and the row and column for $0$ still count as part of the table.
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