I am a three-digit number
Which prime factor will surely appear among my first factors?
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I am a three-digit number
Which prime factor will surely appear among my first factors?
To solve this problem, we will make use of divisibility rules, particularly focusing on the rule for 2.
Given the options, is the only prime factor that will certainly appear among the first factors of any number ending with 12, as other numbers such as 3, 7, or 11 do not have guaranteed divisibility given non-fixed sum of digits or specific rules not directly applicable.
Therefore, the solution to the problem is .
Write all the factors of the following number: \( 6 \)
Because any number ending in 2 is automatically divisible by 2! Whether it's 112, 312, 512, or 912 - they all have 2 as a factor since they end in an even digit.
These might be factors for some numbers ending in 12, but not all of them. For example, 112 ÷ 7 = 16 (so 7 is a factor), but 212 ÷ 7 = 30.28... (so 7 is not a factor).
Super easy! Just look at the last digit. If it's 0, 2, 4, 6, or 8 (even numbers), then the whole number is divisible by 2.
No! Since we don't know what the first digit is (the ? could be 1, 2, 3, etc.), we can't guarantee divisibility by any other prime. Only the divisibility rule for 2 works here.
It means the prime factors when you break down the number. For example, 12 = 2 × 2 × 3, so the prime factors are 2 and 3. Since any ?12 number is even, 2 will always appear in its prime factorization.
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