Opsimath

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*.

### 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.

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