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Number Theory Level 4

A positive integer M is divisible by all the number from 11 - 30 inclusive except for two consecutive integers. What is their sum?


The answer is 33.

Jonathan Hsu by Jonathan Hsu , 19, USA

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1 solution

Andy Hayes
Sep 11, 2015

First we list the numbers between 11 11 and 30 30 inclusive:

11 , 12 , 13 , 14 , 15 , 16 , 17 , 18 , 19 , 20 , 21 , 22 , 23 , 24 , 25 , 26 , 27 , 28 , 29 , 30 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30

If M M is not divisible by 11 11 , then it will also not be divisible by 22 22 . Therefore, 11 11 cannot be one of the two numbers. We can eliminate 12 , 13 , 14 , 12, 13, 14, and 15 15 using similar logic. The eliminated numbers are marked in red.

11 , 12 , 13 , 14 , 15 , 16 , 17 , 18 , 19 , 20 , 21 , 22 , 23 , 24 , 25 , 26 , 27 , 28 , 29 , 30 \color{#D61F06}{11}, \color{#D61F06}{12}, \color{#D61F06}{13}, \color{#D61F06}{14}, \color{#D61F06}{15}, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30

We can now establish that M M must be divisible by those eliminated numbers. Thus, the M M must also be divisible by any number that divides the least common multiple of 11 , 12 , 13 , 14 , 11, 12, 13, 14, and 15 15 . We can eliminate 20 , 21 , 22 , 26 , 28 20, 21, 22, 26, 28 , and 30 30 this way.

11 , 12 , 13 , 14 , 15 , 16 , 17 , 18 , 19 , 20 , 21 , 22 , 23 , 24 , 25 , 26 , 27 , 28 , 29 , 30 \color{#D61F06}{11}, \color{#D61F06}{12}, \color{#D61F06}{13}, \color{#D61F06}{14}, \color{#D61F06}{15}, 16, 17, 18, 19, \color{#D61F06}{20}, \color{#D61F06}{21}, \color{#D61F06}{22}, 23, 24, 25, \color{#D61F06}{26}, 27, \color{#D61F06}{28}, 29, \color{#D61F06}{30}

The two numbers must be consecutive. Therefore, we can eliminate 27 27 and 29 29 .

11 , 12 , 13 , 14 , 15 , 16 , 17 , 18 , 19 , 20 , 21 , 22 , 23 , 24 , 25 , 26 , 27 , 28 , 29 , 30 \color{#D61F06}{11}, \color{#D61F06}{12}, \color{#D61F06}{13}, \color{#D61F06}{14}, \color{#D61F06}{15}, 16, 17, 18, 19, \color{#D61F06}{20}, \color{#D61F06}{21}, \color{#D61F06}{22}, 23, 24, 25, \color{#D61F06}{26}, \color{#D61F06}{27}, \color{#D61F06}{28}, \color{#D61F06}{29}, \color{#D61F06}{30}

Now we can eliminate 18 18 because it is a factor of the least common multiple of the eliminated numbers. We can subsequently eliminate 19 19 because the two numbers must be consecutive.

11 , 12 , 13 , 14 , 15 , 16 , 17 , 18 , 19 , 20 , 21 , 22 , 23 , 24 , 25 , 26 , 27 , 28 , 29 , 30 \color{#D61F06}{11}, \color{#D61F06}{12}, \color{#D61F06}{13}, \color{#D61F06}{14}, \color{#D61F06}{15}, 16, 17, \color{#D61F06}{18}, \color{#D61F06}{19}, \color{#D61F06}{20}, \color{#D61F06}{21}, \color{#D61F06}{22}, 23, 24, 25, \color{#D61F06}{26}, \color{#D61F06}{27}, \color{#D61F06}{28}, \color{#D61F06}{29}, \color{#D61F06}{30}

If M M were divisible by 16 16 , then it would also have to be divisible by 24 24 . If this were true, then there would be no more consecutive numbers left for us to choose from. Therefore, the two numbers must be 16 16 and 17 17 , and their sum is 33 \boxed{33} .

4 Helpful 0 Interesting 0 Brilliant 0 Confused

This is a good solution. However, there is a much simpler way to solve it. You can see that if you have two consecutive integers, one of the is going to be even. 16 is the even number because it is 2 4 2^{4} . No other positive integers has these factors. The next smallest number that has 2 4 2^{4} is 32, and 32 is not smaller than 30. Therefore, the first number is 16, and we can see that since 17 is prime, it is the second number. 16+17=33.

Jonathan Hsu - 5 years, 9 months ago

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