AMC 10 · 2013 · #23

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 number's two other-base numerals are read as ordinary numbers and added. Count the numbers whose sum matches a target's last digits.

Pick an answer.

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

Last two digits is a remainder condition.

S ≡ 2N (mod 100)
2STEP 2

Name the last two digits in each base

Each base's last two digits get names.

(10b+a)+(10d+c)≡ 2N (mod 100)
3STEP 3

Units column forces the two base-units digits equal

The units column makes both base digits equal.

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

Tens column becomes a condition on k

The tens column becomes a small condition.

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

Multiplying the counts gives 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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