 Opsimath

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2021/09/10阅读：61主题：默认主题

# Mathematics Interview Questions (XI)

### 51. Sketch and on the same axis.

Solution. They are all odd functions and increasing on . is the point of inflection for both of them. Notice that, when or , ; when or , . You can view here for a sample.

### 52. What would the two sides of a rectangle ( and ) be to maximise the area if , where is a constant?

Solution. From the AM-GM inequality, the area with equality when . Since , when the area is maximised, , .

### 53. Can 1000003 be written as the sum of 2 square numbers?

Solution. No. Suppose there exists such that . Consider mod 4 on both sides, since square numbers appear. Since , but , so there is no solution to this equation.

### 54. Show that when you square an odd number, you always get one more than a multiple of 8.

Solution. All the odd numbers have the form where . Squaring gives . If is even, then is even, so is also even. If is odd, then is also odd, so is still even. This means we can write in the form where .

### 55. Prove that equals infinity.

Solution.

which diverges, so the original series (commonly called the harmonic series) also diverges.

### 56. Prove that for , when , .

Solution. Taking natural logarithms on both sides gives , i.e., . Suppose . Using quotient rule, which is less than 0 when . Therefore, when , . This gives , completing the proof.

### 57. Prove that is irrational.

Solution. Proof by contradiction. Assume that is rational, so it can be written as where and . Squaring both sides gives , so . This means must be a multiple of 3, so is also a multiple of 3. (Otherwise, if is not a multiple of 3, then is not a multiple of 3.) We can then write as for some . Therefore, , i.e., . This means must be a multiple of 3, so is also a multiple of 3, by the same logic above. and are both even, but ; contradiction.

### 58. What are the possible unit digits for perfect squares?

Solution. For , since , we can see that .

### 59. What are the possible remainders when a cube is divided by 9?

Solution. For , since , we can see that .

### 60. Prove that 801,279,386,104 can't be written as the sum of 3 cubes.

Solution. Suppose there exists such that . The sum of digits of 801,279,386,104 is 4, so . However, since , , so there is no solution to this equation. V1