Divisibility Intuition Math Example 4

Follow the full solution, then compare it with the other examples linked below.

Example 4

medium
A number NN leaves remainder 22 when divided by 55 and remainder 11 when divided by 33. What are the possible last digits of NN, and is NN divisible by 1515?

Solution

  1. 1
    Remainder 22 mod 55: NN ends in 22 or 77 (since 12,17,22,27,…12, 17, 22, 27, \ldots all leave remainder 22).
  2. 2
    Remainder 11 mod 33: digit sum ≑1(mod3)\equiv 1 \pmod{3} (not divisible by 33).
  3. 3
    For NN to be divisible by 1515, it must be divisible by both 33 and 55. But it is not divisible by 55 (remainder 22) or 33 (remainder 11), so NN is definitely not divisible by 1515.

Answer

NN ends in 22 or 77; NN is not divisible by 1515.
Divisibility by a product like 15=3Γ—515 = 3 \times 5 requires divisibility by all prime factors simultaneously. A number with any non-zero remainder mod 33 or mod 55 cannot be divisible by 1515.

About Divisibility Intuition

Understanding when one whole number divides evenly into another, leaving no remainderβ€”the foundation of factor and multiple relationships.

Learn more about Divisibility Intuition β†’

More Divisibility Intuition Examples