Solved on Nov 07, 2023

Explain: Think of an even number. Multiply it by $5$ . What is the last digit of the $5x$ product? Explain why the last digit is always $0$ or $5$ .

STEP 1

Assumptions1. We are dealing with an even number.. The even number is a positive integer.
3. We are multiplying the even number by5.
4. We are interested in the last digit of the product.

STEP 2

Let's denote the even number as $n$ .

STEP 3

Since $n$ is an even number, it can be written in the form $2k$ , where $k$ is an integer.

STEP 4

We are asked to multiply the even number by. So, the product will be $n$ or $(2k)$ , which simplifies to $10k$ .

STEP 5

The last digit of a number in decimal representation is determined by its remainder when divided by10.

STEP 6

Since $10k$ is a multiple of10, its remainder when divided by10 is0.

STEP 7

Therefore, the last digit of the product of an even number and5 is always0.
The last digit of the product of an even number and5 is0.

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Explain: Think of an even number. Multiply it by $5$ . What is the last digit of the $5x$ product? Explain why the last digit is always $0$ or $5$ .

STEP 1

Assumptions1. We are dealing with an even number.. The even number is a positive integer.
3. We are multiplying the even number by5.
4. We are interested in the last digit of the product.

STEP 2

Let's denote the even number as $n$ .

STEP 3

Since $n$ is an even number, it can be written in the form $2k$ , where $k$ is an integer.

STEP 4

We are asked to multiply the even number by. So, the product will be $n$ or $(2k)$ , which simplifies to $10k$ .

STEP 5

The last digit of a number in decimal representation is determined by its remainder when divided by10.

STEP 6

Since $10k$ is a multiple of10, its remainder when divided by10 is0.

STEP 7

Therefore, the last digit of the product of an even number and5 is always0.
The last digit of the product of an even number and5 is0.

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Explain: Think of an even number. Multiply it by 555. What is the last digit of the 5x5x5x product? Explain why the last digit is always 000 or 555.

STEP 1

Assumptions1. We are dealing with an even number.. The even number is a positive integer.3. We are multiplying the even number by5.4. We are interested in the last digit of the product.

STEP 2

Let's denote the even number as nnn.

STEP 3

Since nnn is an even number, it can be written in the form 2k2k2k, where kkk is an integer.

STEP 4

We are asked to multiply the even number by. So, the product will be nnn or (2k)(2k)(2k), which simplifies to 10k10k10k.

STEP 5

The last digit of a number in decimal representation is determined by its remainder when divided by10.

STEP 6

Since 10k10k10k is a multiple of10, its remainder when divided by10 is0.

STEP 7

Therefore, the last digit of the product of an even number and5 is always0.The last digit of the product of an even number and5 is0.

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

Explain: Think of an even number. Multiply it by $5$ . What is the last digit of the $5x$ product? Explain why the last digit is always $0$ or $5$ .

Assumptions1. We are dealing with an even number.. The even number is a positive integer.
3. We are multiplying the even number by5.
4. We are interested in the last digit of the product.

Let's denote the even number as $n$ .

Since $n$ is an even number, it can be written in the form $2k$ , where $k$ is an integer.

We are asked to multiply the even number by. So, the product will be $n$ or $(2k)$ , which simplifies to $10k$ .

Since $10k$ is a multiple of10, its remainder when divided by10 is0.

Therefore, the last digit of the product of an even number and5 is always0.
The last digit of the product of an even number and5 is0.