You note that $3^{16} \equiv 1 \pmod{64}$ and $3^{11} \equiv 59 \pmod{64}$.asif e elahi wrote:How $y \equiv 11 \pmod{16}$ ?harrypham wrote: Therefore $3^y \equiv 59 \pmod{64}$. It follows that $y \equiv 11 \pmod{16}$. Hence $3^y \equiv 3^{11} \equiv 7 \pmod{17}$.
Search found 12 matches
- Tue Sep 10, 2013 12:16 pm
- Forum: Number Theory
- Topic: $2^{x}=3^{y}+5$
- Replies: 4
- Views: 3442
Re: $2^{x}=3^{y}+5$
- Mon Sep 09, 2013 12:36 pm
- Forum: National Math Camp
- Topic: Equation!
- Replies: 5
- Views: 7606
Re: Equation!
The equation is equivalent to $|\sqrt{x-1}-2|+|\sqrt{x-1}-3|=1$.SANZEED wrote:Find all real solution of \[\sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+8-6\sqrt{x-1}}=1\].
From here we use the inequality $|a|+|b| \ge |a+b|$.
- Mon Sep 09, 2013 12:01 pm
- Forum: Number Theory
- Topic: $2^{x}=3^{y}+5$
- Replies: 4
- Views: 3442
Re: $2^{x}=3^{y}+5$
More stronger: $2^x=3^y+5^z$.
- Mon Sep 09, 2013 11:59 am
- Forum: Number Theory
- Topic: $2^{x}=3^{y}+5$
- Replies: 4
- Views: 3442
Re: $2^{x}=3^{y}+5$
Find all positive integers x,y such that $2^{x}=3^{y}+5$ It is easy to see that $(x,y)=(3,1),(5,3)$. If $x \ge 6$ then $64|2^x$. Therefore $3^y \equiv 59 \pmod{64}$. It follows that $y \equiv 11 \pmod{16}$. Hence $3^y \equiv 3^{11} \equiv 7 \pmod{17}$. Since $3^y+5 \equiv 2 \pmod{3}$ then $2^x \equ...
- Sun Aug 11, 2013 9:06 pm
- Forum: Number Theory
- Topic: modulo
- Replies: 1
- Views: 2073
Re: modulo
We have $m|a-b$ with $m,a,b \in \mathbb{Z}, \; m \ne 0$ then we write $a \equiv b \pmod{m}$. Some property: $a \equiv b \pmod{m}, \; b \equiv c \pmod{m}$ then $a \equiv c \pmod{m}$. $a \equiv c \pmod{m}, \; b \equiv d \pmod{m}$ then $ab \equiv cd \pmod{m}$. $a \equiv b \pmod{m}$ then $a+c \equiv b+c...
- Sat Aug 10, 2013 7:35 am
- Forum: Number Theory
- Topic: Can you find an integer b such that:3b^+3b+7?
- Replies: 2
- Views: 2663
Re: Can you find an integer b such that:3b^+3b+7?
From here we obtain $b|7$.liyuqingru wrote:Find $b$ such that \[3b^2+3b+7\vdots b\]
Who can help me solve this question in detail? Thanks a lot!
- Sat Aug 10, 2013 7:30 am
- Forum: Number Theory
- Topic: Pairs of integers
- Replies: 2
- Views: 2958
Re: Pairs of integers
We use Vieta Jumping to solve this problem.SANZEED wrote:Find infinitely many pairs of integers $(a,b)$ such that $1<a<b$ and $ab$ divides $a^{2}+b^{2}-1$. Also find all positive integers $k$ such that there exists $(a,b)$ such that $\frac {a^{2}+b^{2}-1}{ab}=k$.
- Sat Aug 10, 2013 7:26 am
- Forum: Secondary: Solved
- Topic: Dhaka Secondary 2010/12
- Replies: 5
- Views: 12240
Re: Dhaka Secondary 2010/12
Could you please show me your solution ? Thank you.leonardo shawon wrote:answer: 213
- Sat Aug 10, 2013 7:21 am
- Forum: Secondary Level
- Topic: Solve it!
- Replies: 5
- Views: 4479
Re: Solve it!
If $x \ge 2$ then $4|y^2+1$. Therefore $y^2 \equiv 3 \pmod{4}$, a contradiction.Phlembac Adib Hasan wrote:Find all non-negative integer $x,y$ such that $y^2+1=2^x$
(It's a mathlinks problem!I solved yesterday.)
Thus, $x=1$ or $x=0$.
- Thu Aug 08, 2013 12:06 pm
- Forum: Secondary Level
- Topic: Download books
- Replies: 3
- Views: 14036