AMC 10 · 2013 · #18
Grade 6 countingThe number 2013 has the property that its units digit is the sum of its other digits, that is 2+0+1=3. How many integers less than 2013 but greater than 1000 have this property?
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: A four-digit number has the special property that its units digit equals the sum of its other three digits (for example 2013, since 2+0+1=3). Count how many whole numbers strictly between 1000 and 2013 have this property.
Givens: The property: units digit = (thousands digit) + (hundreds digit) + (tens digit); The number must be greater than 1000 and less than 2013, so it runs from 1001 to 2012; Every place holds a single digit from 0 to 9
Unknowns: How many numbers from 1001 to 2012 satisfy units digit = sum of the other three digits
Understand
Restated: A four-digit number has the special property that its units digit equals the sum of its other three digits (for example 2013, since 2+0+1=3). Count how many whole numbers strictly between 1000 and 2013 have this property.
Givens: The property: units digit = (thousands digit) + (hundreds digit) + (tens digit); The number must be greater than 1000 and less than 2013, so it runs from 1001 to 2012; Every place holds a single digit from 0 to 9
Plan
Primary tool: #2 Make a Systematic List
Secondary: #4 Introduce a Variable, #7 Identify Subproblems
Name the four digits so the property becomes one clean equation, then note the units digit can be at most 9 to turn it into a size limit on the other digits. The thousands digit is only 1 or 2, so I split into two subproblems and count each with an organized list: for thousands digit 1 the job is counting digit pairs whose sum is small enough, and for thousands digit 2 only the numbers 2000-2012 are in range, so I just check them.
Execute — Answer: D
4.NBT.A.2 Step 1 Name the four digits
- Write the number as a thousands digit a, hundreds digit b, tens digit c, and units digit d.
- The property says the units digit equals the sum of the other three, so d = a + b + c.
- Because the number is between 1001 and 2012, the thousands digit a is either 1 or 2.
💡 Giving each place its own letter turns a wordy digit rule into a single equation you can work with.
6.EE.B.8 Step 2 The units digit caps the sum
- The units digit d is a single digit, so it can be at most 9.
- Since d = a + b + c, the sum of the other three digits can be at most 9.
- This is the key limit: a + b + c <= 9.
💡 There is no room past 9 in a single digit, so the three other digits must add up to 9 or less.
6.EE.B.5 Step 3 Count the case a = 1
- For thousands digit a = 1 (numbers 1001 to 1999), the limit becomes 1 + b + c <= 9, that is b + c <= 8.
- Count the digit pairs (b, c) with b + c <= 8.
- Group them by the value of the sum s = b + c: if s = 0 there is 1 pair, if s = 1 there are 2 pairs, and so on up to s = 8 which has 9 pairs.
- Adding these gives 1 + 2 + ...
- + 9 = 45.
- Each pair fixes d = 1 + b + c, so it makes exactly one valid number, and every such number lies between 1001 and 1999.
💡 Sorting the pairs by their sum turns the count into a neat triangular addition.
6.EE.B.5 Step 4 Count the case a = 2
- For thousands digit a = 2 the number must stay under 2013, so it is one of 2000 to 2012.
- Here the hundreds digit b = 0, and the property needs d = 2 + 0 + c = 2 + c.
- For 2000-2009 the tens digit c = 0, needing d = 2, which gives 2002 (a valid hit).
- For 2010-2012 the tens digit c = 1, needing d = 3, but the units digit only reaches 2 in this range, so nothing works.
- So exactly 1 number, 2002, comes from this case.
💡 The 2000s barely reach 2012, so only the one number 2002 can satisfy the rule.
6.EE.B.5 Step 5 Add the two cases
- Combine the counts from the two subproblems: 45 numbers from the 1000s plus 1 number from the 2000s.
- That gives 45 + 1 = 46 numbers with the property.
- So the answer is (D).
💡 The two cases never overlap, so the total is just their counts added together.
4.NBT.A.2 Write the number as a thousands digit a, hundreds digit b, tens digit c, and uni 6.EE.B.8 The units digit d is a single digit, so it can be at most 9. Since d = a + b + c 6.EE.B.5 For thousands digit a = 1 (numbers 1001 to 1999), the limit becomes 1 + b + c <= 6.EE.B.5 For thousands digit a = 2 the number must stay under 2013, so it is one of 2000 6.EE.B.5 Combine the counts from the two subproblems: 45 numbers from the 1000s plus 1 nu Review
Reasonableness: The 1000s case alone gives 45, and the 2000s add just one more (2002), landing at 46, which matches choice (D). Choice (C) 45 is the trap for anyone who forgets the lone 2002; note that 2013 itself is excluded because we need numbers strictly below 2013, so it never gets counted. The count 45 = 1+2+...+9 also makes sense: as the hundreds digit drops from 8 down to 0, each step allows one more valid tens digit.
Alternative: Count the 1000s case by pattern instead of by sums: fix the hundreds digit b, then c can be any digit with c <= 8 - b, giving 9 - b choices. As b runs 8, 7, ..., 0 the choices are 1, 2, ..., 9, again totaling 45. Then add the single number 2002 for 46.
CCSS standards used (min grade 6)
4.NBT.A.2Read and write multi-digit whole numbers and compare using symbols (Breaking the four-digit number into its thousands, hundreds, tens, and units digits and reading the range 1001 to 2012 by place value)6.EE.B.8Write an inequality of the form x > c or x < c and graph on a number line (Turning 'the units digit is a single digit' into the size limit a + b + c <= 9, then b + c <= 8)6.EE.B.5Understand solving an equation or inequality as a process of finding values (Counting the digit pairs (b, c) that satisfy b + c <= 8 and checking which 2000-2012 numbers satisfy d = 2 + c)
⭐ Since the units digit maxes out at 9, the other three digits must add to 9 or less; count those digit pairs for the 1000s (45 of them) and add the single 2000s number 2002 to get 46.
⭐ Since the units digit maxes out at 9, the other three digits must add to 9 or less; count those digit pairs for the 1000s (45 of them) and add the single 2000s number 2002 to get 46.
More like this
Same archetype — closest grade level first.