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?
We need to determine which prime factor is guaranteed for the number's appearance. The number is a three-digit number ending with 5, indicated as .
Based on divisibility rules, a number ending in 5 is divisible by 5. Therefore, 5 must be a factor.
Thus, the prime factor that surely appears in the factorization of any number ending in 5 is .
Therefore, the solution to the problem is .
Write all the factors of the following number: \( 6 \)
This comes from the divisibility rule for 5: any number ending in 0 or 5 is divisible by 5. Since your number ends in 5, it must have 5 as a factor, regardless of what the first two digits are!
Absolutely! The number could have other prime factors like 3, 7, 11, etc. But the question asks which prime factor will surely appear, meaning it's guaranteed for every possible number.
Perfect examples! 115 = 5 × 23 and 225 = 5² × 3². Both contain 5 as a prime factor, proving that any three-digit number ending in 5 will have 5 in its prime factorization.
For 5, it's simple: if the last digit is 0 or 5, the number is divisible by 5. For 2, the last digit must be even. For 3, add all digits - if that sum is divisible by 3, so is the original number!
Great question! 5 could appear multiple times in the factorization. For example, 125 = 5³, so 5 appears three times. But it will appear at least once in any number ending in 5.
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