Tag: maths

$ Let\quad x=a(a+1)(a+2)(a+3)\ Now\quad whether\quad a\quad is\quad even\quad or\quad odd,\quad the\quad four\quad consecutive\quad number's\quad product\quad \ contain\quad 2,\quad 3,\quad 4\quad as\quad factors.\ \therefore \quad x\quad is\quad divisible\quad by\quad 2\times 3\times 4=24----(1)\ Option\quad A\longrightarrow Four\quad consecutive\quad numbers\quad contain\quad two\quad even\quad numbers.\ Therefore\quad product\quad is\quad even.\therefore \quad x+1\quad should\quad be\quad odd.\ Option\quad A\quad is\quad correct.\ Option\quad B\longrightarrow x\quad is\quad divisible\quad by\quad 24\quad (from\quad 1)\ Now\quad there\quad exists\quad no\quad prime\quad of\quad the\quad form\quad 24p+1\quad where\quad p\quad is\quad a\quad natural\quad number.\ e.g.\quad \quad 29=24\times 1+5\ \quad \quad \quad \quad \quad 31=24\times 1+7\ \quad \quad \quad \quad \quad 37=24\times 1+13\ With\quad increasing\quad count\quad the\quad second\quad term\quad increases\quad because\quad difference\quad between\ primes\quad increase\quad with\quad higher\quad count.\quad Therefore\quad x+1=n\quad is\quad not\quad a\quad prime\quad under\quad the\quad \ given\quad condition.\quad For\quad a\quad expanded\quad number\quad to\quad be\quad square,\quad the\quad first\quad and\quad last\quad term\ should\quad be\quad square\quad number.   Option\quad C\longrightarrow x=a(a+1)(a+2)(a+3)\ when\quad a\quad is\quad even\quad a=2p\ \therefore \quad x=2p(2a+1)(2a+2)(2a+3)\ \quad \quad \quad =4p(2p+1)(p+1)(2p+3)\ The\quad product\quad is\quad 4\times 1\times 1\times 3=12\ If\quad we\quad add\quad 1\quad then\quad last\quad term\quad of\quad the\quad product\quad is\quad 13\ which\quad is\quad not\quad a\quad square\quad number.\ So\quad x+1=n\quad is\quad not\quad a\quad square\quad number\quad when\quad a\quad is\quad even.\ (2)\quad When\quad a\quad is\quad odd-\ x=(2p+1)(2p+2)(2p+3)(2p+4)\ The\quad last\quad term\quad of\quad the\quad product\quad is\quad 1\times 2\times 3\times 4=24.\ 24+1=25\quad is\quad a\quad square\quad term.\ \therefore \quad n=x+1\quad is\quad a\quad square\quad number\quad when\quad a\quad is\quad odd.\ \therefore \quad Option\quad C\quad is\quad correct\quad when\quad a\quad is\quad odd.\ Option\quad D\longrightarrow \quad Obviously\quad option\quad D\quad is\quad not\quad correct. $

We have,

When the number is divided by $14$ it gives a remainder of $8$,

The number $= 14N + 8 (14N$ is divisible by $14)$

When same number is divided by $7$ it will give remainder $1.$

number should be of the form $4k+1$

$\begin{matrix} \dfrac { { 20 } }{ { 100 } } \times 170000=20\times 1700 \ =34000\, \, \, Ans. \  \end{matrix}$

$Speed\>of\>train\>=\>60\>(\frac{Km}{hr})\\=(\frac{60\cdot\>1000\>m}{3600\>sec})\\=(\frac{50}{3})m/sec\\length\>crossed\>in\>9\>seconds\>=\>(\frac{50}{3})\cdot\>9\>=\>150\>m\\\therefore\>length\>of\>train\>=\>150m$