AMC 10 · 2018 · #19

Grade 7 probability

Archetype Units-Digit Tracking └ Sub-archetype Power Cycle Lookup

units-digit-trackingmodular-arithmetic-mod-10pattern-recognitionprobability-basic pattern-recognitioncasework ↑ Prerequisites: units-digit-trackingprobability-basic

📏 Medium solution 💡 3 insights

Problem

Pick a number m at random from {11,13,15,17,19} and a number n at random from the 20 integers {1999,2000,…,2018}. Find the probability that mⁿ ends in the digit 1.

Pick an answer.

(A)

$\frac{1}{5}$

(B)

$\frac{1}{4}$

(C)

$\frac{3}{10}$

(D)

$\frac{7}{20}$

(E)

$\frac{2}{5}$

AMC 10 2018 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Look for a Pattern

Tool #5 (Look for a Pattern): the units digit of a power repeats in a short cycle, so instead of computing giant powers we just track that repeating cycle for each base's last digit. Tool #9 (Solve an Easier Related Problem): the giant numbers 11¹⁹⁹⁹ etc. shrink to a question about single last digits. Tool #7 (Identify Subproblems): split the problem into five separate cases, one for each value of m, and find the chance of ending in 1 in each. Tool #2 (Make a Systematic List): list the cycle for each last digit and count how many of the 20 exponents land on a 1.

1STEP 1

Only last digits matter

The units digit of mⁿ depends only on m's last digit, so replace the bases with their last digits 1,3,5,7,9 and study each.

11,13,15,17,19 → last digits 1,3,5,7,9

2STEP 2

Find each cycle

Cycle each last digit: 1 stays 1; 3 and 7 hit 1 only when the exponent is a multiple of 4; 9 hits 1 on even exponents; 5 never hits 1.

3ⁿ:3,9,7,1 | 7ⁿ:7,9,3,1 | 9ⁿ:9,1

3STEP 3

Count winning exponents in 1999..2018

In 1999–2018 there are 5 multiples of 4 and 10 evens, so last digit 1 wins 20, 3 wins 5, 5 wins 0, 7 wins 5, and 9 wins 10.

#{mult. of 4}=5, #{even}=10

4STEP 4

Add up the favorable pairs

The 100 pairs are equally likely; winning counts sum to 20+5+0+5+10=40, so the probability is $\frac{40}{100}$ = $\frac{2}{5}$ , choice (E).

\frac{20+5+0+5+10}{100}

\frac{40}{100}

\frac{2}{5}

→ (E)

Answer

\frac{2}{5}

The five bases contribute fractions

\frac{20}{20}

\frac{5}{20}

\frac{0}{20}

\frac{5}{20}

\frac{10}{20}

, i.e. 1,

\frac{1}{4}

,0,

\frac{1}{4}

\frac{1}{2}

. Averaging these five equally likely cases gives

\frac{1}{5}

(1+

\frac{1}{4}

+0+

\frac{1}{4}

\frac{1}{2}

\frac{1}{5}

· 2=

\frac{2}{5}

, matching

\frac{40}{100}

. The value

\frac{2}{5}

=0.4 sits sensibly between 0 and 1, and it should beat the naive

\frac{1}{5}

guess because the base ending in 1 always wins.

💡Key takeaway

Last digits of powers run in short repeating loops, so count how many exponents land on a 1 for each base, add them up, and divide by all 100 pairs to get $\frac{2}{5}$ , choice (E).

Only last digits matter
Find each cycle
Count winning exponents in 1999..2018
Add up the favorable pairs

Only last digits matter

Find each cycle

Count winning exponents in 1999..2018

Add up the favorable pairs

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