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Opsimath

2021/09/18  阅读:19  主题:默认主题

Mathematics Interview Questions (XVIII)

91. Suggest prime factors of 612612503503.

Solution. Note the pattern of that number: 612612503503=503+503000+612000000+612000000000. Hence, 612612503503=1001*(503+612000000)=7*11*13*612000503. As a result, 7, 11, and 13 are all prime factors of it. Also, 612000503 itself is a prime, which can be verified by some cumbersome computation.

Extension. There are actually some other ways of doing this; in particular, if one did not notice the factor 1001, they may try direct divisions by 7, 11, and 13. What is the pattern of the number if it is divisible by 7, 11, and 13, then?

92. How many faces are there on an icosahedron?

Solution. 20. The prefix icosa- (which comes from Ancient Greek εἴκοσι (eíkosi)) means literally 'twenty'.

Extension. From 1-10, the prefixes are mono-, di-, tri-, tetra-, penta-, hexa-, hepta-, octa-, enna-, and deca-. From 11-20, they are hendeca-, dodeca-, trideca, tetradeca-, ...

93. Integrate .

Solution. Letting gives , so . Also, . This implies

Extension. Consider another way of doing it (which is much simpler). Show the equivalence of the two answers.

94. What is the greatest value of for which is divisible of ?

Solution. The factors of which contain 2 are only, where has two 2's, has three, has four, and the others all contain one only. Thus, the greatest value of is .

Extension. What about changing 20 to where is an arbitary natural number?

95. Prove that the product of four consecutive integers is divisible by 24.

Solution. Since , it is sufficient to show that one of the 4 numbers is divisible by 3, one of them is divisible by 4, and another is divisible by 2. This is obvious.

Extension. Generalize this argument. Can we still argue in the same way if we change four into ?

Opsimath

2021/09/18  阅读:19  主题:默认主题

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Opsimath