AMC 10 · 2024 · #6

Grade 8 number-theory

Archetype Base-Representation Digit Count └ Sub-archetype Power-of-Base Extraction

base-conversionprime-factorizationexponentsdigit-counting easier-related-problembound-inequality-then-enumerateidentify-subproblems ↑ Prerequisites: exponentsplace-valuebase-conversion

📏 Medium solution 💡 3 insights

Problem

The number 5×10¹³ is written out as a numeral in base 5 instead of base 10. Count how many digits that base-5 numeral has. The problem offers log₁₀ 5 ≈ 0.7 as a good enough approximation.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

How to solve

Strategy Solve an Easier Related Problem

Converting a fourteen-digit number to base 5 by hand sounds hopeless, until you notice that 5×10¹³ is secretly built out of fives: 10 = 2 × 5. Rewriting it as 2¹³ × 5¹⁴ turns the giant number into a small number times a power of 5, and multiplying by a power of 5 only parks zeros on the end of a base-5 numeral. The whole job then shrinks to counting the base-5 digits of one four-digit number, which takes two lines. Naming the digit count d first makes 'how many digits' precise as a bracket between two powers of 5, so no logarithm is needed at all.

1STEP 1

Turn digit-counting into powers

Digits come from sandwiching between powers of five.

5^d-1 ≤ N ＜ 5^d

2STEP 2

Split the tens into 2s and 5s

Ten to the thirteenth is two and five, each to the thirteenth.

N = 5 × 10¹³, 10¹³ = (2 × 5)¹³ = 2¹³ × 5¹³

3STEP 3

Gather every factor of 5

It becomes 8192 times five to the fourteenth.

N = 5 × 2¹³ × 5¹³ = 2¹³ × 5¹⁴ = 8192 × 5¹⁴

4STEP 4

A power of 5 only adds zeros

That power just appends fourteen zeros.

d = (base-5 digits of 8192) + 14

5STEP 5

Sandwich 8192

8192 sits between the fifth and sixth powers, so six digits.

5⁵ = 3125 ≤ 8192 ＜ 15625 = 5⁶

6STEP 6

Add the shift back on

Six plus fourteen is 20 digits.

d = 6 + 14 = 20, 5¹⁹ ≤ 5 × 10¹³ ＜ 5²⁰

Answer

Check the bracket against real numbers: 5¹⁹ = 19,073,486,328,125 ≈ 1.9×10¹³ and 5²⁰ = 95,367,431,640,625 ≈ 9.5×10¹³, and 5×10¹³ does sit between them, so 20 places is right. The numeral itself confirms it: 8192 in base 5 is 230232, since 2(3125)+3(625)+0(125)+2(25)+3(5)+2 = 8192, so 5×10¹³ in base 5 is 230232 followed by 14 zeros, which is 6 + 14 = 20 digits. A rough size check agrees too: each base-5 digit carries only about 0.7 of a decimal digit's worth of information, so a number of size 10¹3.7 needs roughly 13.7/0.7 ≈ 19.6, and the digit count has to be the next whole number up, 20. That matches choice (B).

💡Key takeaway

To count digits in base 5, find the two neighbouring powers of 5 the number sits between; the exponent of the smaller one plus 1 is the digit count.

Turn digit-counting into powers
Split the tens into 2s and 5s
Gather every factor of 5
A power of 5 only adds zeros
Sandwich 8192 between powers of 5
Add the shift back on