Complete the Number: Make 62_ Even by Divisibility

Q: Why does only 0 work from the given choices?

From the options given (0, 1, 3, 5), only 0 is even. Since divisibility by 2 requires the last digit to be even, 0 is the only choice that makes $ 620 $ divisible by 2.

Q: Do I need to do the actual division to check?

No! You can use the divisibility rule as a shortcut. If the last digit is even, the entire number is divisible by 2. But doing $ 620 \div 2 = 310 $ is a good way to double-check.

Q: What other numbers would make this work?

Any even digit: $ 622 $, $ 624 $, $ 626 $, or $ 628 $ would all be divisible by 2. The pattern is 62[even digit].

Q: Why can't odd numbers ever work?

Because odd numbers always end in 1, 3, 5, 7, or 9. These digits make the entire number odd, and odd numbers are never divisible by 2 without a remainder.

Divisibility Rules with Even Digit Placement

Complete the numbers to obtain a number divisible by 2 without remainder.

$62\text{ }_{—\text{ }}$

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations

00:00 Complete the digit so that the number is divisible by 2

00:03 A number where the units digit is even is divisible by 2

00:07 According to this method, we will go through all the numbers and eliminate accordingly

00:26 And this is the solution to the question

Step-by-step written solution

Follow each step carefully to understand the complete solution

Understand the problem

Complete the numbers to obtain a number divisible by 2 without remainder.

$62\text{ }_{—\text{ }}$

Step-by-step solution

To solve this problem, we identify the numbers that can make $62\text{ }_{—}$ divisible by 2:

A number is divisible by 2 if its last digit is even.
In our choices, the even numbers are $0$ , as it satisfies the divisibility rule for 2.

Substitute the even number:

Try $0$ : The number becomes $620$ .
$620 \div 2 = 310$ , which is an integer showing $620$ is divisible by 2.

Therefore, the correct digit to ensure $62\text{ }_{—}$ is divisible by 2 is 0.

Final Answer

Key Points to Remember

Essential concepts to master this topic

Rule: A number is divisible by 2 if last digit is even
Technique: Check each option: $620 \div 2 = 310$ works perfectly
Check: Verify division gives whole number with no remainder ✓

Common Mistakes

Avoid these frequent errors

Choosing any digit without checking divisibility
Don't just pick random digits like 1, 3, or 5 = odd numbers that make 621, 623, or 625! These are all odd and cannot be divisible by 2. Always choose even digits (0, 2, 4, 6, 8) for the last position.

Practice Quiz

Test your knowledge with interactive questions

Is the number 43 divisible by 4?

FAQ

Everything you need to know about this question

Why does only 0 work from the given choices?

From the options given (0, 1, 3, 5), only 0 is even. Since divisibility by 2 requires the last digit to be even, 0 is the only choice that makes $620$ divisible by 2.

What if there were other even numbers in the choices?

Any even digit would work! If the choices included 2, 4, 6, or 8, those would also be correct answers. The key is that the last digit must be even.

Do I need to do the actual division to check?

No! You can use the divisibility rule as a shortcut. If the last digit is even, the entire number is divisible by 2. But doing $620 \div 2 = 310$ is a good way to double-check.

What other numbers would make this work?

Any even digit: $622$ , $624$ , $626$ , or $628$ would all be divisible by 2. The pattern is 62[even digit].

Why can't odd numbers ever work?

Because odd numbers always end in 1, 3, 5, 7, or 9. These digits make the entire number odd, and odd numbers are never divisible by 2 without a remainder.

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