# Thread: 4 Inequality and number base problems

1. ## 4 Inequality and number base problems

The following math problems consists of two number base problems and two inequality problems that I have been unable to solve:

1. A number X is converted into base 7 and becomes a four-digit number. Its leftmost digit is removed and written again as the rightmost digit. The number thus obtained in base 7 is twice X. Find the decimal representations of all such numbers X.

2. Does there exist a positive integer which, when it is written in base 10 and its leftmost digit is crossed out, can be multiplied by 56 to get the original number?

3. Two real numbers, a and b satisfy the equation ab = 1.
(i) Prove that a6 + 4b6 >/= 4.
(ii) Does the inequality a6 +4b6 > 4 hold for all a and b such that ab = 1?

4. Numbers a, b and c are all greater than or equal to 0. Prove
2(a3 + b3 + c3) >/= ab(a+b) + bc(b+c) + ca(c + a)

Thanks a lot for answers to anyone of these questions

2. ## Re: 4 Inequality and number base problems

Hello, JeanMal!

1. A number X is converted into base 7 and becomes a four-digit number.
Its leftmost digit is removed and written again as the rightmost digit.
The number thus obtained in base 7 is twice X.
Find the decimal representations of all such numbers X.

This is an alphametic in base-7:. $\begin{array}{cccc} A&B&C&D \\ \times &&& 2 \\ \hline B&C&D&A \end{array}$

I found two solutions: . $\begin{array}{cccc}1&2&5&4 \\ \times &&& 2 \\ \hline 2&5&4&1 \end{array}\;\;\text{ and }\;\;\begin{array}{cccc}2&5&4&1 \\ \times &&& 2 \\ \hline 5&4&1&2 \end{array}$

Therefore: . $X \;=\;\begin{Bmatrix}1254_7 &=& 480 \\ 2541_7 &=& 960 \end{Bmatrix}$

3. ## Re: 4 Inequality and number base problems

Hello, JeanMal!

2. Does there exist a positive integer which, when it is written in base 10
and its leftmost digit is crossed out, can be multiplied by 56 to get the original number?

Let $a$ be the leftmost digit.
Let $b$ be an $n$-digit integer.
. . Then the number has the form: . $N \:=\:10^na + b$

Crossing out the leftmost digit, we have $b.$
And we want $56b$ to equal $N.$

We have: . $56b \:=\:10^na + b \quad\Rightarrow\quad 55b \:=\:10^na$

. . . . . . . . $5\!\cdot\!11b \:=\:2\!\cdot\!5\!\cdot\!10^{n-1}a \quad\Rightarrow\quad11b \:=\: 2\!\cdot\!10^{n-1}a$

The equation holds if: . $b = 2\!\cdot\!10^{n-1}\;\text{ and }\;a = 11$

But $a = 11$ is not a digit.

4. ## Re: 4 Inequality and number base problems

Originally Posted by JeanMal

3. Two real numbers, a and b satisfy the equation ab = 1.
(i) Prove that a6 + 4b6 >/= 4.
(ii) Does the inequality a6 +4b6 > 4 hold for all a and b such that ab = 1?
We have $b=\displaystyle\frac{1}{a}$.
Replace b in the inequality: $a^6+\frac{4}{a^6}\ge 4$.
Multiply both members by $a^6$: $a^{12}+4\ge 4a^6\Leftrightarrow \left(a^6-2\right)^2\ge 0$
The equality holds for $a=\displaystyle\sqrt[6]{2}$.
Then the second inequality isn't true for $a=\sqrt[6]{2}, \ b=\displaystyle\frac{1}{\sqrt[6]{2}}$

5. ## Re: 4 Inequality and number base problems

Originally Posted by JeanMal

4. Numbers a, b and c are all greater than or equal to 0. Prove
2(a3 + b3 + c3) >/= ab(a+b) + bc(b+c) + ca(c + a)
Use the inequality $a^2-ab+b^2\ge ab$ and multiply both members with $(a+b)$.
We get $a^3+b^3\ge ab(a+b)$.
In the same way $b^3+c^3\ge bc(b+c)$ and $c^3+a^3\ge ca(c+a)$.