# What is the least integer n such that 1/2^n < 0.001

What is the least integer \$n\$ such that \$1/2^n<0.001\$?

1. \$10\$
2. \$11\$
3. \$500\$
4. \$501\$
5. There is no such least integer.

So, you were trying to be a good test taker and practice for the GRE with PowerPrep online. Buuuut then you had some questions about the quant section—specifically question 12 of the second Quantitative section of Practice Test 1. Those questions testing our knowledge of Exponents and Roots can be kind of tricky, but never fear, PrepScholar has got your back!

## Survey the Question

Let’s search the problem for clues as to what it will be testing, as this will help shift our minds to think about what type of math knowledge we’ll use to solve this question. Pay attention to any words that sound math-specific and anything special about what the numbers look like, and mark them on our paper.

We see that \$n\$ is an exponent in our question, so we’ll likely use our Exponents and Roots math skill. And since we have an inequality in our question, we’ll likely use our Solving Linear Inequalities math skill. Let’s keep what we’ve learned about these skills at the tip of our minds as we approach this question.

## What Do We Know?

Let’s carefully read through the question and make a list of the things that we know.

1. We want to find the least integer value of \$n\$
2. We know that: \$1/2^n<0.001\$

## Develop a Plan

We want to find the least value of \$n\$, so let’s try to simplify the inequality in our question until it’s easier to see what values \$n\$ can have.

## Solve the Question

When solving an equation or inequality for a variable, it’s usually best to rearrange the equation until we have that variable in the numerator of any fractions, isolated by itself. Looking at our inequality, let’s start by multiplying both sides of the inequality by \$2^n\$ to remove it from the denominator of the fraction. Anytime we multiply or divide an inequality by a number, we need to double check to make sure that number is positive. If it is negative, we need to reverse the inequality’s sign direction. We know that \$2\$ raised to ANY exponent will give us a positive result, so we don’t need to worry about reversing the inequality’s sign direction here.

 \$1/2^n\$ \$<\$ \$0.001\$ \$ \$ \$ \$ \$1\$ \$<\$ \$0.001·2^n\$

Let’s continue isolating the \$n\$ by dividing both sides of the inequality by \$0.001\$.

 \$1\$ \$<\$ \$0.001·2^n\$ \$ \$ \$ \$ \$1/0.001\$ \$<\$ \$2^n\$ \$ \$ \$ \$ \$1{,}000\$ \$<\$ \$2^n\$

Ah, this is much easier to understand! So \$2^n\$ must be greater than \$1{,}000\$. Although there technically is a mathematical way to solve for \$n\$, that is definitely beyond the scope of the GRE. So instead, let’s use our scrap paper to make a table of different \$n\$ and \$2^n\$ values, along with whether or not \$2^n\$ is greater than \$1{,}000\$. So let’s start with \$n=1\$ and use our math knowledge or GRE calculator to keep multiplying by \$2\$.

 \$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;n\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\$ \$2^n\$ Is \$2^n\$ greater than \$1{,}000\$? \$1\$ \$2\$ No \$2\$ \$4\$ No \$3\$ \$8\$ No \$4\$ \$16\$ No \$5\$ \$32\$ No \$6\$ \$64\$ No \$7\$ \$128\$ No \$8\$ \$256\$ No \$9\$ \$512\$ No \$10\$ \$1{,}024\$ Yes

Nicely done! so it looks like \$n=10\$ is the first integer value for \$n\$ where \$2^n\$ is greater than \$1{,}000\$. The correct answer is A, \$10\$.

## What Did We Learn

Simplifying the inequality was definitely very helpful. It changed a very abstract looking inequality and turned it into an easier to understand concept: finding a value for \$n\$ where \$2^n\$ would be greater than \$1{,}000\$. So a great lesson to learn is simplify whenever possible.

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