AMC 10 · 2013 · #25
Grade 6 number-theoryBernardo chooses a three-digit positive integer N and writes both its base-5 and base-6 representations on a blackboard. Later LeRoy sees the two numbers Bernardo has written. Treating the two numbers as base-10 integers, he adds them to obtain an integer S. For example, if N=749, Bernardo writes the numbers 10,444 and 3,245, and LeRoy obtains the sum S=13,689. For how many choices of N are the two rightmost digits of S, in order, the same as those of 2N?
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 three-digit number N (from 100 to 999) is written in base five and in base six. Each of those two numerals is then read as an ordinary base-ten number, and the two are added to get S. Count how many N make the last two digits of S equal to the last two digits of 2N.
Givens: N is a three-digit integer, so 100 <= N <= 999.; N5 is the base-five numeral of N read as a base-ten number; N6 is the base-six numeral read as a base-ten number.; S = N5 + N6.; We compare the last two digits of S with the last two digits of 2N.
Unknowns: How many of the 900 possible values of N make the last two digits of S match the last two digits of 2N.
Understand
Restated: A three-digit number N (from 100 to 999) is written in base five and in base six. Each of those two numerals is then read as an ordinary base-ten number, and the two are added to get S. Count how many N make the last two digits of S equal to the last two digits of 2N.
Givens: N is a three-digit integer, so 100 <= N <= 999.; N5 is the base-five numeral of N read as a base-ten number; N6 is the base-six numeral read as a base-ten number.; S = N5 + N6.; We compare the last two digits of S with the last two digits of 2N.
Plan
Primary tool: #4 Introduce a Variable
Secondary: #8 Analyze the Units, #7 Identify Subproblems, #3 Eliminate Possibilities
The phrase 'last two digits' is a signal to work mod 100, and mod 100 keeps only the last two digits of each numeral. So the first move is to name those digits with variables: a and b for base five, c and d for base six. That turns a scary base-conversion question into one congruence. Then split the congruence into two easier subproblems, the units column (mod 10) and the tens column, solve each, and finally count the surviving cases.
Execute — Answer: E
5.NBT.A.1 Step 1 Turn 'last two digits' into mod 100
- Two numbers share their last two digits exactly when they are equal mod 100, so the target is S congruent to 2N modulo 100.
- Reading a base-five numeral as a base-ten number only respaces its digits; taking the result mod 100 keeps just the last two of them.
- So only the last two base-five digits of N and the last two base-six digits can matter.
💡 Mod 100 erases every column except tens and units, so only the last two digits of each numeral survive.
4.NBT.B.6 Step 2 Name the last two digits in each base
- Split N by division.
- In base five the units digit is a = N mod 5 and the fives digit is b = floor(N/5) mod 5, so N5 ends in the block 'b a', worth 10b + a mod 100.
- In base six the units digit is c = N mod 6 and the sixes digit is d = floor(N/6) mod 6, worth 10d + c.
- The condition becomes (10b + a) + (10d + c) congruent to 2N modulo 100.
💡 Dividing by 5 and by 6 hands you the exact digits that land in the ones and tens columns.
6.EE.A.2 Step 3 Units column forces the two base-units digits equal
- Look mod 10 first.
- Since a is at most 4 and c is at most 5, the sum a + c is at most 9, so the units digit of S is just a + c.
- Also 2N mod 10 equals 2 times (N mod 5) = 2a.
- Matching the units gives a + c congruent to 2a modulo 10, so c congruent to a modulo 10.
- Since a and c both lie between 0 and 5, this forces c = a.
- Then N leaves the same remainder a (with a at most 4) mod 5 and mod 6, so N is congruent to a modulo 30.
- That leaves 5 possible unit residues a = 0, 1, 2, 3, 4.
💡 The ones column can only balance when the two base-remainders are literally the same number.
5.NBT.A.2 Step 4 Tens column becomes a condition on k
- Write N = 30k + a with a in {0, 1, 2, 3, 4}.
- Then 2N = 60k + 2a, and since c = a the condition (10b + a) + (10d + a) congruent to 60k + 2a collapses to 10(b + d) congruent to 60k modulo 100, that is b + d congruent to 6k modulo 10.
- Dividing N shows b = floor(N/5) mod 5 = 6k mod 5 = k mod 5, and d = floor(N/6) mod 6 = 5k mod 6.
- So the whole problem is now: for which k is (k mod 5) + (5k mod 6) congruent to 6k modulo 10?
💡 Stripping away the matched units column leaves a single clean congruence in k alone.
4.NBT.B.5 Step 5 Count the good k, then multiply
- The test (k mod 5) + (5k mod 6) congruent to 6k modulo 10 depends only on k mod 30.
- Checking one full period, it holds exactly for k congruent to 0, 1, 12, 13, 24 modulo 30 - five residues.
- For N = 30k + a to be three-digit with a up to 4, k runs from 4 to 33, which is 30 consecutive values, so each residue mod 30 occurs exactly once.
- That yields 5 valid k (namely 12, 13, 24, 30, 31), and each pairs with 5 choices of a.
- So the count is 5 times 5 = 25, which is (E).
💡 One clean congruence gives five k-blocks, and each block carries five units, so the total is a tidy 5-by-5 grid.
5.NBT.A.1 Two numbers share their last two digits exactly when they are equal mod 100, so 4.NBT.B.6 Split N by division. In base five the units digit is a = N mod 5 and the fives d 6.EE.A.2 Look mod 10 first. Since a is at most 4 and c is at most 5, the sum a + c is at 5.NBT.A.2 Write N = 30k + a with a in {0, 1, 2, 3, 4}. Then 2N = 60k + 2a, and since c = a 4.NBT.B.5 The test (k mod 5) + (5k mod 6) congruent to 6k modulo 10 depends only on k mod Review
Reasonableness: The 25 solutions fall into five neat blocks of five consecutive integers: 360-364, 390-394, 720-724, 900-904, and 930-934. Each block starts at a multiple of 30 (k = 12, 13, 24, 30, 31) and runs through the units a = 0..4, matching the 5-by-5 structure. Spot-check N = 360: base five 2420 gives last two digits 20, base six 1400 gives 00, sum ends in 20; 2N = 720 also ends in 20, so they agree. The count 25 is the largest choice, fitting a problem tuned so every valid block sits fully inside the three-digit range.
Alternative: Skip the algebra and brute-force it: for each N from 100 to 999 build its base-five and base-six numerals, read them as base-ten numbers, add, and compare the last two digits with those of 2N. Counting the hits gives 25 directly. The algebra just explains why the winners cluster into five blocks of five.
CCSS standards used (min grade 6)
5.NBT.A.1Recognize that a digit in one place represents ten times as much as to its right (Reading the last two digits of a base-five or base-six numeral as 10 times the tens digit plus the units digit, and seeing that mod 100 keeps only those two.)4.NBT.B.6Find whole-number quotients and remainders with up to four-digit dividends (Extracting the base digits a = N mod 5, b = floor(N/5) mod 5, c = N mod 6, d = floor(N/6) mod 6 by division with remainder.)6.EE.A.2Write, read, and evaluate expressions in which letters stand for numbers (Setting up and reducing the congruence in a, b, c, d, k to derive c = a and the condition on k.)5.NBT.A.2Explain patterns in the number of zeros and placement of the decimal point (Using the times-10 place-value shift to turn 10(b + d) congruent to 60k into b + d congruent to 6k modulo 10.)4.NBT.B.5Multiply a whole number of up to four digits by a one-digit whole number (Combining 5 valid values of k with 5 values of a to get the final count 5 times 5 = 25.)
⭐ 'Last two digits' means work mod 100: line up the ones column first, then the tens column, and a scary base-conversion problem turns into simple remainder counting.
⭐ 'Last two digits' means work mod 100: line up the ones column first, then the tens column, and a scary base-conversion problem turns into simple remainder counting.
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