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Number Theory

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152
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There are three consecutive positive integers. The smallest one is a multiple of 3, the middle one is a multiple of 5 and the largest one is a multiple of 4. What is the smallest possible sum of those three integers?

Nov 1, 2022

2+0 Answers

#1
+1624
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The middle number is divisible by 5 so it must end in 0 or 5. If it ends in 0, then the next number ends in a 1, however the problem stated it was divisible by 4, and we know that the number must be even, so the last number ends in a 6. The middle number ends in a 5, and the first number ends in a 4. If the first number ends in a 4, we can consider 2 digit numbers, so 24 is div by 3, 25 is div by 5, but 26 is not div by 4. I can say 24 is div by 3 because of the divisibility rule for 3 (the sum of the digits of the number is div by 3), Lets try (2+3) and 4, so 54, 55, and 56. We can see those are divisible by 3, 5, then 4. 54 + 56 + 55 = 110 + 55 = 165

Nov 1, 2022
#2
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Incorrect answer.

Guest Nov 7, 2022