The number of integers n that satisfy the inequalities $|n -60| < |n-100| <| n-20|$ is

Before diving into the solution, let's understand what these absolute value expressions represent:

$|n-60|$ = distance from point $n$ to point $60$ on the number line

$|n-100|$ = distance from point $n$ to point $100$ on the number line

$|n-20|$ = distance from point $n$ to point $20$ on the number line

Think of it this way: If you're standing at position $n$ on a number line, these expressions tell you how far you are from the points 60, 100, and 20 respectively.

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We need to solve: $|n-60| < |n-100| < |n-20|$

This gives us two separate conditions:

1. $|n-60| < |n-100|$ (n is closer to 60 than to 100)
2. $|n-100| < |n-20|$ (n is closer to 100 than to 20)

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When is a point closer to 60 than to 100?

The answer lies in finding the midpoint between 60 and 100:

$\text{Midpoint} = \frac{60 + 100}{2} = \frac{160}{2} = 80$

Any point to the left of 80 is closer to 60, and any point to the right of 80 is closer to 100.

Let's verify:

At $n = 80$: $|80-60| = 20$ and $|80-100| = 20$ (equal distances)

At $n = 79$: $|79-60| = 19$ and $|79-100| = 21$ (closer to 60)

At $n = 81$: $|81-60| = 21$ and $|81-100| = 19$ (closer to 100)

Therefore: $|n-60| < |n-100|$ when $n < 80$

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When is a point closer to 100 than to 20?

We find the midpoint between 100 and 20:

$\text{Midpoint} = \frac{100 + 20}{2} = \frac{120}{2} = 60$

Let's verify:

At $n = 60$: $|60-100| = 40$ and $|60-20| = 40$ (equal distances)

At $n = 61$: $|61-100| = 39$ and $|61-20| = 41$ (closer to 100)

At $n = 59$: $|59-100| = 41$ and $|59-20| = 39$ (closer to 20)

Therefore: $|n-100| < |n-20|$ when $n > 60$

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From our analysis:

Condition 1: $n < 80$

Condition 2: $n > 60$

Combined: $60 < n < 80$

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The integers satisfying $60 < n < 80$ are:

$n \in \{61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79\}$

Counting: From 61 to 79 inclusive = $79 - 61 + 1 = 19$ integers

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Let's verify with $n = 70$:

$|70-60| = 10$

$|70-100| = 30$

$|70-20| = 50$

Check: $10 < 30 < 50$

Answer: There are 19 integers that satisfy the given inequalities.

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When comparing distances using absolute values, the midpoint between two reference points is the boundary where distances become equal. This technique works for any similar absolute value inequality problem.