Number Play | Class 8 Mathematics Notes

Checking Divisibility Quickly

This section revisits and explains shortcuts to check divisibility by numbers such as 2, 4, 5, 8, and 10, using algebraic reasoning based on place values. It begins by expressing numbers in the Indian number system as sums of place values, for example, a 5-digit number as e×10000 + d×1000 + c×100 + b×10 + a.

The section explains why divisibility by 10 depends solely on the units digit being zero, since all other place values are multiples of 10. Similar algebraic reasoning is encouraged for divisibility by 5, 2, 4, and 8.

The section then introduces a shortcut for divisibility by 9, explaining that any number made up of digits 0 and 9 is divisible by 9 because each term in its expanded form is a multiple of 9. However, this shortcut alone does not identify all multiples of 9.

Students observe that the remainder when multiples of 10 or 100 are divided by 9 corresponds to the digit in the tens or hundreds place, respectively. This leads to the method of adding digits repeatedly to find the remainder upon division by 9.

The section explains that a number is divisible by 9 if and only if the sum of its digits is divisible by 9. It also introduces the concept of digital roots and their relation to divisibility by 9.

This section develops students' understanding of divisibility tests and prepares them for shortcuts for other numbers like 3 and 11.

📊 Diagram: See figure_13: ? Is 10 divisible by 9? If not, what is the remainder?; See figure_14 and figure_15: Visual explanation of remainders when dividing by 9; See figure_16: 10000 = 9999 + 1, and so on.

🧪 Activity: Students practice checking divisibility by 9 using digit sums and digital roots, and explore remainders for multiples of 10 and 100.

🔗 Connection: This section introduces divisibility shortcuts that lead to the next sections on divisibility by 3 and 11.