वह छोटी से छोटी संख्या क्या होगी जिसे 5 तथा 6 तथा 7 तथा 8 से भाग देने पर 3 शेष बताओ परंतु 9 से भाग देने पर 0 शेष बचता हो?
Answers
Answer:
wah chhoti sankhya jisme 5,6,7,8 se bhag dene par 3 shesh bache = 5,6,7,8 ka LCM + 3
5= 1x5
6 = 2x3
7 = 1x7
8 = 2x2x2
LCM = 2x2x2x3x5x7 = 840
isliye number = 840 + 3 = 843
number 9 se bhajya honi chahiye
843, 9 se bhajya nahi hai
isliye vanchhit sankya aisi honi chahiye jo do sharto ko pura kar sake
1. wo 840*a + 3 ke rup me honi chahiye, jahan a koi poornank hai.
2. wo 9 se bhajya honi chahiye
(iske aage ke liye kripya sabse niche diya note padhe)
840a + 3 = 0 (mod 9)
840a = -3 (mod 9)
840a = 6 (mod 9)
3a = 6 (mod 9)
a= 2(mod 3)
a = 2 + 3n
chunki sankhya sabse chhoti honi chahiye
isliye n = 0 (yahan yah mana ja raha hai ki sankhya dhanatmak honi chahiye athwa rinatmak sankhya lene par iska koi Uttar nahi hoga kyunki wo rinatmak anant tak jayega)
a= 2
sankhya = 840x2 + 3
= 1683
(note: agar ye prashn 10 ya usse niche class ka hai to LCM nikalne ke baad hame trial and error method se solution nikalna chahiye arthat 840a+3 me a= 1,2 rakh kar dekhna chahiye ki kaun si sankhya sahi hai)