AMC 10 · 2013 · #25

Grade 6 number-theory
Archetype Small-Domain Constrained Search └ Sub-archetype Factor / Divisibility Constraint
base-conversionmodular-arithmeticplace-value identify-subproblemscasework ↑ Prerequisites: base-conversionmodular-arithmetic
📏 Medium solution 💡 3 insights
Problem
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.

Pick an answer.

(A)
5
(B)
10
(C)
15
(D)
20
(E)
25

AMC 10 2013 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Introduce a Variable

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.

1STEP 1

Turn 'last two digits' into mod 100

Two numbers share their last two digits exactly when they agree mod 100, and mod 100 keeps only the last two digits of each numeral.

S ≡ 2N (mod 100)
2STEP 2

Name the last two digits in each base

Name the digits: in base five a = N mod 5 and b = N/5 rounded down, mod 5; in base six c = N mod 6 and d = N/6 rounded down, mod 6.

(10b+a)+(10d+c)≡ 2N (mod 100)
3STEP 3

Units column forces the two base-units digits equal

Since a + c is at most 9, the ones column reads a + c ≡ 2a (mod 10), forcing c = a, so N ≡ a (mod 30) with a in 0 to 4.

a+c≡ 2a (mod 10) → c=a, N≡ a (mod 30)
4STEP 4

Tens column becomes a condition on k

With N = 30k + a the tens column collapses to b + d ≡ 6k (mod 10), where b = k mod 5 and d = 5k mod 6.

10(b+d)≡ 60k (mod 100) → b+d≡ 6k (mod 10), b=k mod 5, d=5k mod 6
5STEP 5

Count the good k, then multiply

It holds for k ≡ 0, 1, 12, 13, 24 (mod 30); k runs over the 30 values 4 to 33, so 5 good k times 5 values of a = 25, choice (E).

5 valid k × 5 values of a = 25
Answer
25
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.
💡Key takeaway

'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.

  • Turn 'last two digits' into mod 100
  • Name the last two digits in each base
  • Units column forces the two base-units digits equal
  • Tens column becomes a condition on k
  • Count the good k, then multiply

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