Similarly, explain using algebra why the divisibility shortcuts for 5, 2, 4, and 8 work.

Question

Homi · Accepted Answer

We can use algebra to represent any number and understand divisibility rules.

Step 1 — Divisibility by 5

Any number is a multiple of 10 plus its last digit. Let $N$ be any whole number. Let $d_0$ be its last digit. Let $K$ be the number formed by remaining digits. For $N = 123$ , $d_0 = 3$ and $K = 12$ . We can write $N$ as:

$N = K \times 10 + d_0$

We want to check if $N$ is divisible by 5. This means $K \times 10 + d_0$ must be divisible by 5. We know that $10$ is divisible by 5. So, $K \times 10$ is always divisible by 5. For $N$ to be divisible by 5, $d_0$ must be divisible by 5. Possible values for $d_0$ are 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. The only digits divisible by 5 are 0 and 5.

$\boxed{\text{A number is divisible by 5 if its last digit is 0 or 5.}}$

Step 2 — Divisibility by 2

Any number is a multiple of 10 plus its last digit. Let $N$ be any whole number. Let $d_0$ be its last digit. Let $K$ be the number formed by remaining digits. We can write $N$ as:

$N = K \times 10 + d_0$

We want to check if $N$ is divisible by 2. This means $K \times 10 + d_0$ must be divisible by 2. We know that $10$ is divisible by 2. So, $K \times 10$ is always divisible by 2. For $N$ to be divisible by 2, $d_0$ must be divisible by 2. Possible values for $d_0$ are 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. The digits divisible by 2 are 0, 2, 4, 6, 8.

$\boxed{\text{A number is divisible by 2 if its last digit is 0, 2, 4, 6, or 8.}}$

Step 3 — Divisibility by 4

Any number is a multiple of 100 plus its last two digits. Let $N$ be any whole number. Let $L_2$ be the number formed by its last two digits. For $N = 12345$ , $L_2 = 45$ . Let $K$ be the number formed by digits before the last two. For $N = 12345$ , $K = 123$ . We can write $N$ as:

$N = K \times 100 + L_2$

We want to check if $N$ is divisible by 4. This means $K \times 100 + L_2$ must be divisible by 4. We know that $100$ is divisible by 4.

$100 = 4 \times 25$

So, $K \times 100$ is always divisible by 4. For $N$ to be divisible by 4, $L_2$ must also be divisible by 4.

$\boxed{\text{A number is divisible by 4 if its last two digits form a number divisible by 4.}}$

Step 4 — Divisibility by 8

Any number is a multiple of 1000 plus its last three digits. Let $N$ be any whole number. Let $L_3$ be the number formed by its last three digits. For $N = 12345$ , $L_3 = 345$ . Let $K$ be the number formed by digits before the last three. For $N = 12345$ , $K = 12$ . We can write $N$ as:

$N = K \times 1000 + L_3$

We want to check if $N$ is divisible by 8. This means $K \times 1000 + L_3$ must be divisible by 8. We know that $1000$ is divisible by 8.

$1000 = 8 \times 125$

So, $K \times 1000$ is always divisible by 8. For $N$ to be divisible by 8, $L_3$ must also be divisible by 8.

$\boxed{\text{A number is divisible by 8 if its last three digits form a number divisible by 8.}}$

Answer

(i) A number is divisible by 5 if its last digit is 0 or 5. (ii) A number is divisible by 2 if its last digit is 0, 2, 4, 6, or 8. (iii) A number is divisible by 4 if the number formed by its last two digits is divisible by 4. (iv) A number is divisible by 8 if the number formed by its last three digits is divisible by 8.

Question 21

Answer

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