dear sir W, Mjs the set {1,4,n} have the condition that if two different elements are selected and 2112 is added to the result , then the result is a perfect square if n is a positif number . then the number of possible values of n is (A) 8 (B) 7 (C) 6 (D) 5 (E) 4


Question and Answers Forum

All Questions Topic List
Algebra Questions
Previous in All Question Next in All Question
Previous in Algebra Next in Algebra
Question Number 78399 by john santu last updated on 17/Jan/20

$d e a r s i r W, M j s$ $t h e s e t {1, 4, n} h a v e t h e c o n d i t i o n t h a t$ $i f t w o d i f f e r e n t e l e m e n t s a r e$ $s e l e c t e d a n d 2112 i s a d d e d t o$ $t h e r e s u l t, t h e n t h e r e s u l t$ $i s a p e r f e c t s q u a r e i f n i s a$ $p o s i t i f n u m b e r . t h e n t h e n u m b e r$ $o f p o s s i b l e v a l u e s o f n i s$ $(A) 8 (B) 7 (C) 6 (D) 5$ $(E) 4$
Commented by mr W last updated on 17/Jan/20

$q u e s t i o n i s n o t c l e a r!$ $w h a t m e a n s ‘ ‘ t w o d i f f e r e n t e l e m e n t s$ $a r e s e l e c t e t a n d 2112 i s a d d e d t o t h e$ ${r e s u l t}^{″} w h e n t w o d i f f e r e n t e l e m e n t s$ $a r e s e l e c t e d, s a y 1 a n d 4, w h a t i s t h e$ $r e s u l t ? t h e s u m o f t h e s e t w o e l e m e n t s$ $o r t h e p r o d u c t o f t h e s e e l e m e n t s o r$ $s o m e t h i n g e l s e ?$ $p l e a s e m a k e t h e q u e s t i o n c l e a r .$
Commented by john santu last updated on 17/Jan/20

$t h e p u r p o s e o f t h i s p r o b l e m$ $i s i f 2 m e m b e r s a r e s e l e c t e d$ $f r o m t h e s e t f o r e x a m p l e$ $1 a n d n t h e n 1 \times n + 2112 t h e$ $r e s u l t a r e q u a d r a t i c n u m b e r$
Commented by mr W last updated on 17/Jan/20

$n o w c l e a r! s o t h e q u e s t i o n s h o u l d b e$ $. . .$ $i f t w o d i f f e r e n t e l e m e n t s a r e$ $s e l e c t e d a n d 2112 i s a d d e d t o$ $t h e i r p r o d u c t, t h e n t h e r e s u l t$ $. . .$
Commented by john santu last updated on 17/Jan/20

$s o h o w m u c h i s t h e v a l u e o f n$ $, s i r ?$
Commented by john santu last updated on 17/Jan/20

$i g o t i n f i n i t y s i r . i s r i g h t ?$
Commented by MJS last updated on 17/Jan/20

$1 \times 4 + 2112 = 2116 = 46^{2}$ $\Rightarrow n can be any number$ $1 \times n + 2112 = p^{2} \Rightarrow infinite solutions for n$ $4 \times n + 2112 = p^{2} \Rightarrow infinite solutions for n$
Commented by john santu last updated on 17/Jan/20

$y e s s i r . i g o t t h e s a m e a n s w e r$
Commented by mr W last updated on 17/Jan/20

$b u t w h a t i f n s h o u l d a l s o b e a p e r f e c t$ $s q u a r e ?$ $1 \times n + 2112 = p^{2}$ $4 \times n + 2112 = p^{2}$
Commented by john santu last updated on 17/Jan/20

$n i s p o s i t i v e n u m b e r s i r$
Commented by mr W last updated on 17/Jan/20

$i k n o w . i f n i s o n l y p o s i t i v e i n t e g e r,$ $t h e r e a r e i n f i n i t e p o s s i b i l i t i e s . b u t$ $i f n s h o u l d a l s o b e p e r f e c t s q u a r e,$ $t h e n i t h i n k t h e p o s s i b i l i t i e s a r e$ $f i n i t e!$
Commented by MJS last updated on 17/Jan/20
$let n=(p−q)^2 (1) (p−q)^2 +2112=p^2 ⇒ p=((q^2 +2112)/(2q)) p, q∈N∧0<q≤p ⇒ 0<q≤45 ⇒ q∈{2, 4, 6, 8, 12, 16, 22, 24, 32, 44} p∈{529, 266, 179, 136, 94, 74, 59, 56, 49, 46} n∈{277729, 68644, 29929, 16384, 6724, 3364, 1369, 1024, 289, 4} (2) 4(p−q)^2 +2112=p^2 ⇒ q=p−((√(p^2 −2112))/2) ⇒ p∈{46, 56, 74, 94, 136, 266} q∈{45, 40, 45, 53, 72, 135} n∈{17161, 4096, 1681, 841, 256, 1} still infinite possibilities for n with 1×4+2112=46^2$
$let n = {(p - q)}^{2}$ $(1) {(p - q)}^{2} + 2112 = p^{2} \Rightarrow p = \frac{q^{2} + 2112}{2 q}$ $p, q \in N \land 0 < q ⩽ p \Rightarrow 0 < q ⩽ 45$ $\Rightarrow$ $q \in {2, 4, 6, 8, 12, 16, 22, 24, 32, 44}$ $p \in {529, 266, 179, 136, 94, 74, 59, 56, 49, 46}$ $n \in {277729, 68644, 29929, 16384, 6724, 3364, 1369, 1024, 289, 4}$ $(2) 4 {(p - q)}^{2} + 2112 = p^{2} \Rightarrow q = p - \frac{\sqrt{p^{2} - 2112}}{2}$ $\Rightarrow$ $p \in {46, 56, 74, 94, 136, 266}$ $q \in {45, 40, 45, 53, 72, 135}$ $n \in {17161, 4096, 1681, 841, 256, 1}$ $still infinite possibilities for n with$ $1 \times 4 + 2112 = 46^{2}$
Commented by Rasheed.Sindhi last updated on 17/Jan/20

$n s h o u l d b e s u c h t h a t$ $(1 \times n + 2112) & (4 \times n + 2112)$ $b o t h a r e p e r f e c t s q u a r e s .$
Commented by john santu last updated on 17/Jan/20

$y e s s i r . w h a t i s v a l u e o f n p o s s i b l e ?$
Commented by john santu last updated on 17/Jan/20

$i n m y b o o k t h e a n s w e r i s 7$
	Answered by john santu last updated on 17/Jan/20

	$1 \times n + 2112 = k^{2} \Rightarrow n = k^{2} - 2112$ $4 \times n + 2112 = p^{2} \Rightarrow n = \frac{p^{2} - 2112}{4}$ $k^{2} - 2112 = \frac{p^{2} - 2112}{4}$ $k^{2} - \frac{p^{2}}{4} = 2640 \Rightarrow (k - \frac{p}{2}) (k + \frac{p}{2}) = 2640$
	Commented by Rasheed.Sindhi last updated on 17/Jan/20

	$B u t s i r n i s p o s i t i v e .$
	Commented by MJS last updated on 17/Jan/20

	$yes . I posted a l l 9 solutions, then I wrote the$ $last 2 lines . . .$
	Commented by MJS last updated on 17/Jan/20
	$error in last line k^2 −(p^2 /4)=1584 your idea is good, it leads to n∈{−512, −87, 97, 697, 3072, 8497, 16113, 37888, 155497} since we′re looking for n∈N we have indeed 7 solutions$
	$error in last line$ $k^{2} - \frac{p^{2}}{4} = 1584$ $your idea is good, it leads to$ $n \in {- 512, - 87, 97, 697, 3072, 8497, 16113, 37888, 155497}$ $since {we}^{'} re looking for n \in N we have indeed$ $7 solutions$
	Commented by john santu last updated on 17/Jan/20

	$w a w . . t h a n k s y o u s i r$
	Answered by Rasheed.Sindhi last updated on 17/Jan/20
	$⇒ { ((1×4+2112=46^2 )),((1×n+2112 is perfect square)),((4×n+2112 is perfect square)) :} 4×n+2112 is perfect square ⇒n+528 is perfect square ⇒ { ((n+2112 is perfect square)),((n+528 is perfect square)) :} Let n+528=q^2 n+2112=n+528+1584=p^2 >q^2 q^2 +1584=p^2 (p+q)(p−q)=1584=2^4 .3^2 .11 p,q∈Z^+ ∧ p>q⇒p+q>p−q •For p,q being whole numbers p+q and p−q are both even. p+q=44⇒p−q=36 p+q=48⇒p−q=33 (rejected) p+q=66⇒p−q=24 p+q=88⇒p−q=18 Continue$
	$\Rightarrow {\begin{cases} 1 \times 4 + 2112 = 46^{2} \\ 1 \times n + 2112 i s p e r f e c t s q u a r e \\ 4 \times n + 2112 i s p e r f e c t s q u a r e \end{cases}$ $4 \times n + 2112 i s p e r f e c t s q u a r e$ $\Rightarrow n + 528 i s p e r f e c t s q u a r e$ $\Rightarrow {\begin{cases} n + 2112 i s p e r f e c t s q u a r e \\ n + 528 i s p e r f e c t s q u a r e \end{cases}$ $L e t n + 528 = q^{2}$ $n + 2112 = n + 528 + 1584 = p^{2} > q^{2}$ $q^{2} + 1584 = p^{2}$ $(p + q) (p - q) = 1584 = 2^{4} . 3^{2} . 11$ $p, q \in Z^{+} \land p > q \Rightarrow p + q > p - q$ $∙ F o r p, q b e i n g w h o l e n u m b e r s$ $p + q a n d p - q a r e b o t h e v e n .$ $p + q = 44 \Rightarrow p - q = 36$ $p + q = 48 \Rightarrow p - q = 33 (r e j e c t e d)$ $p + q = 66 \Rightarrow p - q = 24$ $p + q = 88 \Rightarrow p - q = 18$ $C o n t i n u e$
	Commented by john santu last updated on 17/Jan/20

	$t h a n k s y o u$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum