Search found 7 matches
- Sat Jan 26, 2013 10:23 pm
- Forum: Higher Secondary Level
- Topic: Some Last year Divisional Problems
- Replies: 13
- Views: 10399
Re: Some Last year Divisional Problems
when $p=2$, $p^a(1+p^m) = 2^a(1+2^m)$ here, $1 + 2^m$ is odd and $2^a$ is even as their product is a perfect square, and $1 + 2^m$ cannot be divided by $2^a$ so, both of them are perfect square separately and you can download avro from here- http://www.omicronlab.com/avro-keyboard-download.html in o...
- Sat Jan 26, 2013 1:37 am
- Forum: Geometry
- Topic: Geometry Marathon v1.0
- Replies: 23
- Views: 16467
Re: Geometry Marathon v1.0
চিত্র.jpg সমস্যা ৪- $DBGP$ এবং $PEFC$ ট্রাপিজিয়ামে, $DB = PF, PG = EC$ এবং তাদের চার কোণ সমান, তাই এরা সর্বসম তাই $BF + FG = CG + FG$ , তাই $BF = CG = DP = PE$ এবং $DB = CE = PF = PG$ একইভাবে $AMPE$ এবং $CGPN$ সর্বসম, তাই $PN = PE = AM = CG$ এবং $AN = CE = PM = PG$ একইভাবে $ANPD$ এবং $MPFB$ সর্বসম...
- Fri Jan 25, 2013 8:38 pm
- Forum: Higher Secondary Level
- Topic: Some Last year Divisional Problems
- Replies: 13
- Views: 10399
Re: Some Last year Divisional Problems
Problem 1 - I solved it in a lengthy (also complex) way, I guess Suppose, $a < b$ and $b = a+m$ and $p^a + p^b = k^2$ or, $p^a + p^{a+m} = k^2$ or, $p^a (1 + p^m) = k^2$ Now, we have two subcases, either p is even or odd Subcase 1 - When $p$ is even, $(p=2)$ $1 + 2^m$ cannot be divided by $2$ and t...
- Fri Jan 25, 2013 1:15 am
- Forum: Higher Secondary Level
- Topic: Some Last year Divisional Problems
- Replies: 13
- Views: 10399
Re: Some Last year Divisional Problems
sakib.creza vai, amar kase english english hoileo english mathematical term gula amar kase hebrew, apnake personal msg e bojhanor ekta try nite partam......
- Fri Jan 25, 2013 1:05 am
- Forum: Higher Secondary Level
- Topic: Some Last year Divisional Problems
- Replies: 13
- Views: 10399
Re: Some Last year Divisional Problems
Problem 4 $abbcca = 100001a + 11000b + 110c \equiv 0\pmod {7} .........(1)$ $abc = 100a + 10b + c \equiv 4\pmod{7}$ So, $110000a + 11000b + 1100c \equiv 4\pmod{7}$ , $[\times 1100] ...........(2)$ $(2) - (1)$ $9999a + 990c \equiv 4\pmod{7}$ Or, $99(101a + 10c) \equiv 4\pmod{7}$ So, $99 \times aca \...
- Fri Jan 25, 2013 12:00 am
- Forum: Higher Secondary Level
- Topic: Some Last year Divisional Problems
- Replies: 13
- Views: 10399
Re: Some Last year Divisional Problems
২ নাম্বারটার ক্ষেত্রে, ধরি, সংখ্যাটার মৌলিক উৎপাদকগুলার সেট $\{P_1, P_2, P_3.......P_n\}$ তাহলে, সংখ্যাটার মোট উৎপাদক হবে সেটটার সব উপসেট অর্থাৎ পাওয়ার সেটের উপাদানের সমান ( যেমন উৎপাদকগুলা হবে $P_1 \times P_2, P_1 \times P_2 .......... P_2 \times P_3 \times P_4$.......) অর্থাৎ $2^n$ (যদি মৌলিক উৎপ...
- Fri Dec 21, 2012 10:44 pm
- Forum: Secondary: Solved
- Topic: Dhaka Secondary 2011/6
- Replies: 13
- Views: 19849
Re: Dhaka Secondary 2011/6
shouldn't it look like this?