Now Check Which of These Are Divisible by 11: A Simple Guide

If you’ve ever encountered a number puzzle asking, “Now check which of these are divisible by 11,” you’re likely curious about divisibility rules—especially for 11. Whether you're a student learning math, a teacher creating worksheets, or someone simply loving logic and numbers, understanding how to test divisibility by 11 can be both fun and useful.

In this article, we’ll walk through what it means to be divisible by 11, explore the key divisibility rule, and show you how to quickly check any list of numbers. Let’s dive in!

Understanding the Context

What Does “Divisible by 11” Mean?

A number is divisible by 11 if, when divided by 11, there is no remainder—mathematically expressed as:

A number n is divisible by 11 if n ÷ 11 = k, where k is an integer.

Now check which of these are divisible by 11.

Key Insights

But there’s an even better way to test divisibility by 11 without actual division, especially for quick mental checks: the 11 divisibility rule.

The 11 Divisibility Rule: Step-by-Step

One of the simplest and most effective ways to check if a number is divisible by 11 is the alternating sum rule. Here’s how it works:

  1. Write the number from right to left (units to highest place value).
    For example, if checking 847, write it as 7 (hundreds), 4 (tens), 8 (units) → digits are 8, 4, 7 from left to right, but evaluate right to left:
    Digit places:
    • Position 0 (units): 7
    • Position 1 (tens): 4
    • Position 2 (hundreds): 8

Continue Reading

We reduce $ (y + 1)^{2025} $ modulo $ y^2 + y + 1 $. Since $ y^2 \equiv -y - 1 $, we can reduce all powers of $ y $ to linear expressions.
Let’s compute the reduction of $ (y + 1)^{2025} $ modulo $ y^2 + y + 1 $. Let’s denote $ \omega $ as a primitive cube root of unity (so $ \omega^3 = 1 $, $ \omega
e 1 $, and $ \omega^2 + \omega + 1 = 0 $), and note that this polynomial divides $ y^2 + y + 1 $.

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Final Thoughts

  1. Alternate adding and subtracting digits starting from the rightmost digit.
    Follow this pattern:
    Add: digit at odd position (right to left — starting at 0 → even indices)
    Subtract: digit at even position
    Use the sign pattern: + − + − + (starting from the right-most digit)

  2. Add up the results and see if the total is divisible by 11.
    If the final sum is 0, ±11, ±22, etc., then the original number is divisible by 11.

Example: Is 143 Divisible by 11?

Let’s apply the rule:

  • Number: 143 → read right to left: 3 (pos 0), 4 (pos 1), 1 (pos 2)
  • Compute alternating sum:
    +3 (pos 0) − 4 (pos 1) + 1 (pos 2)
    = 3 − 4 + 1 = 0
  • Since 0 is divisible by 11, 143 is divisible by 11.
    (Indeed, 143 ÷ 11 = 13)

Common Examples You Can Test Now

Try checking these numbers using the rule above:

  • 121 ✅ (1 − 2 + 1 = 0 → divisible)
  • 99 ✅ (9 − 9 = 0 → divisible)
  • 132 ✅ (2 − 3 + 1 = 0 → divisible)
  • 123 ✖ (1 − 2 + 3 = 2 → not divisible)
  • 110 ✅ (0 − 1 + 1 = 0 → divisible)