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Class act (Posted on 2002-05-02) Difficulty: 3 of 5
31 students in a row were numbered 1,2,...,31 in order. The teacher wrote down a number on the blackboard.

Student 1 said "the number is divisible by 1",
Student 2 said "the number is divisible by 2",
and so forth...until
Student 31 said "the number is divisible by 31".

The teacher remarked: "Very well pups, but two of you gave a wrong statement, and those two sit besides each other". Determine those two.

See The Solution Submitted by theBal    
Rating: 3.5000 (10 votes)

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re: Really not that hard | Comment 8 of 15 |
(In reply to Really not that hard by Brian Smith)

The ONLY consecutive pair is 16 and 17? Because in your list I see 2&3, 3&4, 4&5, 7&8, 8&9, and 16&17. I think I know what you are trying to say, but you didn't say it completely. Maybe you meant the largest pair?

The way I figured it out was similar to levik's but with a little more detail. I made a table with the numbers 1-31 down the side, and the prime numbers 2-31 across the top. Then I put the power of each prime number needed in the table. Ex: 24 had a 3 in the two-column, and a 1 in the three-column since 2^3 * 3^1 = 24. The plan is to find two powers I can reduce by one to eliminate ONLY two numbers that happen to be next to each other.

Then I looked for a single greatest power for any one prime number. Ex: the five-column had 1s, and a single 2. The three-column had 1s, 2s, and a single 3. The two-column had 1s, 2s, 3s, and a single 4. So far some of the key numbers to focus on are 16, 25, and 27.

The rest of the prime columns only had 1s in them, but 7, 11, and 13 had more than one 1 which means we can't eliminate them with out eliminating other numbers as well.

So now we are focusing on 16, 17, 19, 23, 25, 27, 29, & 31. NOW we see that there is only one pair of consecutive numbers: 16 and 17. By the way, the number the teacher wrote is 2,123,581,660,200.

  Posted by nikki on 2003-06-01 06:54:38

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