AMC 10 · 2013 · #13
Grade 6 number-theoryHow many three-digit numbers are not divisible by 5, have digits that sum to less than 20, and have the first digit equal to the third digit?
Pick an answer.
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.
Toolkit + CCSS Solution
Understand
Restated: Count the three-digit whole numbers that satisfy all three conditions at once: the number is not a multiple of $5$, the sum of its three digits is less than $20$, and its first (hundreds) digit equals its third (units) digit.
Givens: The number has exactly three digits, so the hundreds digit is at least $1$; The number is not divisible by $5$; The three digits add up to less than $20$; The first (hundreds) digit equals the third (units) digit; Answer choices: (A) $52$, (B) $60$, (C) $66$, (D) $68$, (E) $70$
Unknowns: How many three-digit numbers meet all three conditions
Understand
Restated: Count the three-digit whole numbers that satisfy all three conditions at once: the number is not a multiple of $5$, the sum of its three digits is less than $20$, and its first (hundreds) digit equals its third (units) digit.
Givens: The number has exactly three digits, so the hundreds digit is at least $1$; The number is not divisible by $5$; The three digits add up to less than $20$; The first (hundreds) digit equals the third (units) digit; Answer choices: (A) $52$, (B) $60$, (C) $66$, (D) $68$, (E) $70$
Plan
Primary tool: #2 Make a Systematic List
Secondary: #4 Introduce a Variable, #3 Eliminate Possibilities
The question asks 'how many', so the natural tool is #2 (Make a Systematic List): sort the valid numbers into a few clean cases and count each. To set that up, first use Tool #4 (Introduce a Variable) to name the digits: since the hundreds digit equals the units digit, every number looks like $\overline{aba}$, controlled by just two digits $a$ and $b$. Then Tool #3 (Eliminate Possibilities) throws out the outer digits that break the 'not divisible by $5$' rule, leaving a short list of $a$ values. Finally the digit-sum condition becomes a bound on $b$, and counting the allowed $b$ for each $a$ and adding gives the total.
Execute — Answer: B
2.NBT.A.1 Step 1 Name the digits: the number is aba
- A three-digit number has a hundreds digit, a tens digit, and a units digit.
- The condition 'first digit equals third digit' means the hundreds digit and the units digit are the same.
- Call that shared outer digit $a$ and the middle digit $b$.
- So every number we want has the shape $\overline{aba}$, and it is completely decided by the two digits $a$ and $b$.
💡 Forcing the first and last digits to match turns three free digits into just two, $a$ and $b$.
4.OA.B.4 Step 2 Apply the not-divisible-by-5 rule to a
- A number is divisible by $5$ exactly when its units digit is $0$ or $5$.
- Here the units digit is $a$.
- Since the number has three digits, $a$ cannot be $0$.
- To avoid being divisible by $5$, $a$ also cannot be $5$.
- That leaves the eight values $a \in \{1,2,3,4,6,7,8,9\}$.
- The middle digit $b$ has no divisibility restriction, so for now $b$ can be any digit $0$ through $9$.
💡 Divisibility by 5 is decided entirely by the last digit, and the last digit here is the outer digit $a$.
6.EE.B.5 Step 3 Turn the digit-sum condition into a bound on b
- The three digits are $a$, $b$, and $a$, so their sum is $2a + b$.
- The condition 'digits sum to less than $20$' says $2a + b < 20$, which rearranges to $b < 20 - 2a$.
- For each fixed $a$, this tells us exactly which middle digits $b$ are allowed.
- Remember $b$ is still capped at $9$ because it is a single digit.
💡 The sum condition only limits the middle digit once the outer digit is chosen, so it becomes a ceiling on $b$.
6.EE.B.5 Step 4 Count the allowed b for each a
- Go through each allowed $a$ and count the digits $b$ from $0$ to $9$ with $b < 20 - 2a$.
- When $a$ is $1,2,3,4$, the bound $20 - 2a$ is $18,16,14,12$, all above $9$, so every digit $0$ through $9$ works: $10$ choices each.
- When $a = 6$, $b < 8$ gives $b \in \{0,\dots,7\}$: $8$ choices.
- When $a = 7$, $b < 6$ gives $6$ choices.
- When $a = 8$, $b < 4$ gives $4$ choices.
- When $a = 9$, $b < 2$ gives $2$ choices.
💡 For small outer digits the sum can never reach 20, so nothing is lost; only the big outer digits start cutting choices.
4.OA.A.3 Step 5 Add the cases to get the total
- Add the counts from every case.
- The four values $a = 1,2,3,4$ give $4 \times 10 = 40$ numbers.
- The remaining values give $8 + 6 + 4 + 2 = 20$ numbers.
- Together that is $40 + 20 = 60$ three-digit numbers meeting all three conditions.
- So the answer is $\textbf{(B)}\ 60$.
💡 The cases never overlap and cover everything, so simply summing their counts gives the full total.
2.NBT.A.1 A three-digit number has a hundreds digit, a tens digit, and a units digit. The 4.OA.B.4 A number is divisible by $5$ exactly when its units digit is $0$ or $5$. Here th 6.EE.B.5 The three digits are $a$, $b$, and $a$, so their sum is $2a + b$. The condition 6.EE.B.5 Go through each allowed $a$ and count the digits $b$ from $0$ to $9$ with $b < 2 4.OA.A.3 Add the counts from every case. The four values $a = 1,2,3,4$ give $4 \times 10 Review
Reasonableness: The eight outer digits split into two groups: $a=1,2,3,4$ each allow all $10$ middle digits, and $a=6,7,8,9$ allow $8,6,4,2$. The second group forms an even, decreasing pattern ($8,6,4,2$) that makes sense because a larger outer digit eats more of the budget of $20$, leaving less room for $b$. The maximum possible if no sum condition applied would be $8 \times 10 = 80$ numbers, and our count $60$ is below that, which is expected since the sum condition removes some. $60$ matches choice (B).
Alternative: Complementary counting: without the digit-sum limit there are $8 \times 10 = 80$ numbers of the form $\overline{aba}$ with $a \in \{1,2,3,4,6,7,8,9\}$. Now subtract the ones with $2a + b \ge 20$. This needs $b \ge 20 - 2a$, which is only possible for large $a$: $a=6$ loses $b \in \{8,9\}$ ($2$), $a=7$ loses $b \in \{6,7,8,9\}$ ($4$), $a=8$ loses $6$, $a=9$ loses $8$. That removes $2+4+6+8 = 20$, giving $80 - 20 = 60$, the same answer.
CCSS standards used (min grade 6)
2.NBT.A.1Understand that the three digits of a three-digit number represent amounts of hundreds, tens, and ones (Reading a three-digit number by its hundreds, tens, and units digits so the 'first digit equals third digit' condition becomes the shape aba.)4.OA.B.4Recognize divisibility and multiples of whole numbers (Using the rule that a number is divisible by 5 only when its units digit is 0 or 5 to eliminate a = 0 and a = 5 for the outer digit.)6.EE.B.5Understand solving an inequality as finding which values from a set make it true (Rewriting the digit-sum condition as b < 20 - 2a and counting, for each outer digit a, how many middle digits b satisfy it.)4.OA.A.3Solve multistep problems with whole numbers using the four operations (Combining the per-case counts (four cases of 10, plus 8, 6, 4, 2) into the final total of 60.)
⭐ When the first and last digits must match, the number is really just aba; pin down the outer digit with the divisibility rule, turn the sum condition into a limit on the middle digit, then count case by case.
⭐ When the first and last digits must match, the number is really just aba; pin down the outer digit with the divisibility rule, turn the sum condition into a limit on the middle digit, then count case by case.
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