2021 AMC Fall 10B Problem 19

P16-20~5 minPrint

Exponents, Logarithms & Radicals Bases & Digits Bound the quantity above/below or estimate to pin it down

Let $N$ be the positive integer $7777\ldots777$ , a $313$ -digit number where each digit is a $7$ . Let $f(r)$ be the leading digit of the $r{ }$ th root of $N$ . What is $f(2) + f(3) + f(4) + f(5)+ f(6)?$

Show solution without answering

Key insight

Write N as a front part times a power of 10 that divides evenly by r, then compare the front part with small r-th powers.

Solution

Warm-up. Why does the power of $10$ need to divide evenly? Look at $\sqrt{700}$ and $\sqrt{7000}$ . Writing $700 = 7 \times 10^2$ gives $\sqrt{700} = \sqrt7 \times 10 \approx 26$ , leading digit $2$ . But $7000 = 70 \times 10^2$ gives $\sqrt{7000} = \sqrt{70} \times 10 \approx 84$ , leading digit $8$ . The front part must be chosen so the power of $10$ comes out of the root cleanly.

$N$ has $313$ digits, so $N = 7.777\ldots \times 10^{312}$ . For each root, make the power of $10$ a multiple of $r$ , then find the whole number $d$ with $d^r \le (\text{front part}) < (d + 1)^r$ . That $d$ is the leading digit.

$r = 2$ . $312 = 2 \times 156$ , so the front part is $7.77\ldots$ Since $2^2 = 4 \le 7.77 < 9 = 3^2$ , $f(2) = 2$ .
$r = 3$ . $312 = 3 \times 104$ . Since $1 \le 7.77 < 8 = 2^3$ , $f(3) = 1$ .
$r = 4$ . $312 = 4 \times 78$ . Since $1 \le 7.77 < 16 = 2^4$ , $f(4) = 1$ .
$r = 5$ . $312$ is not a multiple of $5$ . The nearest lower multiple is $310$ , so move the decimal point two places: $N = 777.7\ldots \times 10^{310}$ . Since $3^5 = 243 \le 777.7 < 1024 = 4^5$ , $f(5) = 3$ .
$r = 6$ . $312 = 6 \times 52$ . Since $1 \le 7.77 < 64 = 2^6$ , $f(6) = 1$ .

Sum: $2 + 1 + 1 + 3 + 1 = 8$ .

The answer is $\boxed{\textbf{(A)}\ 8}$ .

Why this works

The leading digit of a root depends only on the front part of the number once the power of $10$ is split off evenly, because the root of $10^{312}$ is just another power of $10$ . This is scientific notation, and it means you never need to know more than the first few digits of $N$ .

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

Since $N$ has $313$ digits, $N = 7.777\ldots \times 10^{312}$ . To take an $r$ th root cleanly, shift the decimal point so the power of $10$ is a multiple of $r$ , then find which integer $d$ satisfies $d^r \le (\text{leading factor}) < (d+1)^r$ .

$r = 2$ : $312 = 2 \cdot 156$ , and $2^2 = 4 \le 7.77 < 9 = 3^2$ , so $f(2) = 2$ .
$r = 3$ : $312 = 3 \cdot 104$ , and $1 \le 7.77 < 8 = 2^3$ , so $f(3) = 1$ .
$r = 4$ : $312 = 4 \cdot 78$ , and $1 \le 7.77 < 16 = 2^4$ , so $f(4) = 1$ .
$r = 5$ : $312$ is not a multiple of $5$ ; write $N = 777.7\ldots \times 10^{310}$ with $310 = 5 \cdot 62$ . Since $3^5 = 243 \le 777.7 < 1024 = 4^5$ , $f(5) = 3$ .
$r = 6$ : $312 = 6 \cdot 52$ , and $1 \le 7.77 < 64 = 2^6$ , so $f(6) = 1$ .

The sum is $2 + 1 + 1 + 3 + 1 = 8$ .

The answer is $\boxed{\textbf{(A)}\ 8}$ .

Review in the AoPS books

Introduction to Algebra, Chapter 1: Follow the Rules (§1.6, §1.7, §1.8)
Introduction to Number Theory, Chapter 8: Base Numbers
Introduction to Algebra, Chapter 11: Special Factorizations (§11.4)
Introduction to Number Theory, Chapter 9: Base Number Arithmetic