1. Given that \[7^{x+7}=8^x \]
and \[x=\log _{b}7^7\]
what is the value of $b$?
2.If $A + B =1, B + C = 2, C + D = 3, ...... X + Y = 25, Y + Z = 26$, find $A – Z$.
3.The equation \[x^3 +3xy + y^3 = 1\]
is solved in integers. Find the possible values of
$x+y$.
Some problems to solve
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Please don't post problems (by starting a topic) in the "X: Solved" forums. Those forums are only for showcasing the problems for the convenience of the users. You can always post the problems in the main Divisional Math Olympiad forum. Later we shall move that topic with proper formatting, and post in the resource section.
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Abdul Muntakim Rafi
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nafistiham
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Re: Some problems to solve
$1.$
given that,
\[7^{x+7}=8^{x}\]
\[\Rightarrow 7^{x}\cdot 7^{7}=8^{x}\]
\[\Rightarrow7^{7}=\left ( \frac{8}{7} \right )^{x}\]
now,\[x=\log_{b}7^7\]
\[\Rightarrow x=\log_{b}\left ( \frac{8}{7} \right )^x\]
\[\Rightarrow\log_{b}\left ( \frac{8}{7} \right )=1\]
\[\Rightarrow\log_{b}\left ( \frac{8}{7} \right )=\log_bb\]
\[\Rightarrow\therefore b=\frac{8}{7}\]
$2.$
here, we just have to subtract the summation of the even equations from the summation of the odd equations.
which means,we'll get
\[A+B-B-C+C+D-D-E+........-Y-Z=-1\]
$3.$
here,
\[x^3+3xy+y^3\]
by trial and error we get that,
$x=2$ and $y=-1$
so,
\[x+y=1\]
given that,
\[7^{x+7}=8^{x}\]
\[\Rightarrow 7^{x}\cdot 7^{7}=8^{x}\]
\[\Rightarrow7^{7}=\left ( \frac{8}{7} \right )^{x}\]
now,\[x=\log_{b}7^7\]
\[\Rightarrow x=\log_{b}\left ( \frac{8}{7} \right )^x\]
\[\Rightarrow\log_{b}\left ( \frac{8}{7} \right )=1\]
\[\Rightarrow\log_{b}\left ( \frac{8}{7} \right )=\log_bb\]
\[\Rightarrow\therefore b=\frac{8}{7}\]
$2.$
here, we just have to subtract the summation of the even equations from the summation of the odd equations.
which means,we'll get
\[A+B-B-C+C+D-D-E+........-Y-Z=-1\]
$3.$
here,
\[x^3+3xy+y^3\]
by trial and error we get that,
$x=2$ and $y=-1$
so,
\[x+y=1\]
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
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Abdul Muntakim Rafi
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Re: Some problems to solve
You are right about the 1st and 3rd problem. 
1.I had a similar approach in 1.
3.Hey,is there any way except using trial and error. I used $a^3+b^3$ formula and showed the equation would be right if $x+y=1$ But don't we have to be sure that there are no other solutions.
2.Is your calculation right?
And is the question right? Cause $A+B=1,B+C=2,.............,Y+Z=25$ . $Y+Z$ should be equal to $25$. but the question says $Y+Z=26$
1.I had a similar approach in 1.
3.Hey,is there any way except using trial and error. I used $a^3+b^3$ formula and showed the equation would be right if $x+y=1$ But don't we have to be sure that there are no other solutions.
2.Is your calculation right?
And is the question right? Cause $A+B=1,B+C=2,.............,Y+Z=25$ . $Y+Z$ should be equal to $25$. but the question says $Y+Z=26$
Man himself is the master of his fate...
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nafistiham
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Re: Some problems to solve
about $2.$
i also tried with $x^3+y^3$ theorem,but could not complete.plse mention the solution.
and, we have to be sure that there is no other solution but $x+y=1$ but, i am not sure how to do it.
$3.$
yap, there is a problem in the question.
if, $A+B=1,B+C=2$ and the sequence go thus, then how can $X+Y$ be $26$?
actually, i didn't see this at the first place. so, i could solve it thinking $X+Y=26$.
but, as it should be $X+Y=25$, i can find out $A+Z$ not $A-Z$.where did you get this question?
i think t would be better to be sure whether it is right or not.i am totally confused.till now i think we cant find out $A-Z$ from the given properties even if $X+Y=25$
i also tried with $x^3+y^3$ theorem,but could not complete.plse mention the solution.
and, we have to be sure that there is no other solution but $x+y=1$ but, i am not sure how to do it.
$3.$
yap, there is a problem in the question.
if, $A+B=1,B+C=2$ and the sequence go thus, then how can $X+Y$ be $26$?
actually, i didn't see this at the first place. so, i could solve it thinking $X+Y=26$.
but, as it should be $X+Y=25$, i can find out $A+Z$ not $A-Z$.where did you get this question?
i think t would be better to be sure whether it is right or not.i am totally confused.till now i think we cant find out $A-Z$ from the given properties even if $X+Y=25$
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
Using $L^AT_EX$ and following the rules of the forum are very easy but really important, too.Please co-operate.
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Abdul Muntakim Rafi
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Re: Some problems to solve
Bogura Higher Secondary 2011... I have tried another way... First assign a value of A... Then see what you get for Z...Then we can get $A-Z$ easily... That gives 13 as an answer if u take $A$ to be $0$. But this gives different values for different values of A. So it's not logical at all.
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nafistiham
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Re: Some problems to solve
here it is not said that the values are integers
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
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Re: Some problems to solve
Rafi, there was a definite prob with no. 2. I did solve it a few days ago, taking y+z=25.
First prob was an easy one. I had the same approach.
3rd prob has another solution. Consider $(x,y)=(-1,-1)$.
I checked all the cases.
First prob was an easy one. I had the same approach.
3rd prob has another solution. Consider $(x,y)=(-1,-1)$.
I checked all the cases.
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nafistiham
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Re: Some problems to solve
@labib
bhya, plz show the solution of $2$ or give a clue
bhya, plz show the solution of $2$ or give a clue
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
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Re: Some problems to solve
Tiham, I solved it for A+z, vai... :p
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Learn how to write equations, and don't forget to read Forum Guide and Rules.
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nafistiham
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Re: Some problems to solve
ohhh!! you just have poured cold water on my dreams.
so, i am expressing the way to get $A+Z$
to sum up all the equations we get
\[A+2B+2C+2D+............+2Y+Z=1+2+3+........+25\]
\[\Rightarrow (A+Z)+2(B+C+D+........Y)=325\]
\[\Rightarrow (A+Z)+2\left \{(B+C)+(D+E)+(F+G)+......(X+Y)\right \}=325\]
\[\Rightarrow (A+Z)+2(2+4+6+........24)=325\]
\[\Rightarrow (A+Z)+312=325\]
\[\Rightarrow (A+Z)=13\]
so, i am expressing the way to get $A+Z$
to sum up all the equations we get
\[A+2B+2C+2D+............+2Y+Z=1+2+3+........+25\]
\[\Rightarrow (A+Z)+2(B+C+D+........Y)=325\]
\[\Rightarrow (A+Z)+2\left \{(B+C)+(D+E)+(F+G)+......(X+Y)\right \}=325\]
\[\Rightarrow (A+Z)+2(2+4+6+........24)=325\]
\[\Rightarrow (A+Z)+312=325\]
\[\Rightarrow (A+Z)=13\]
Last edited by nafistiham on Sat Dec 03, 2011 6:08 pm, edited 1 time in total.
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
Using $L^AT_EX$ and following the rules of the forum are very easy but really important, too.Please co-operate.
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