Math, asked by Hakar, 1 year ago

Only genious (not the rank) can do it :p
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Answers

Answered by MOSFET01
15
\bold{\large{\underline{\underline{Answer}}}}

Consider the number be x

when,

\bold{ \dfrac{x}{2}\: , \: R \: = \: 1}

\bold{ \dfrac{x}{3}\: , \: R \: = \: 2}

\bold{ \dfrac{x}{4}\: , \: R \: = \: 3}

and so on to \bold{ \dfrac{x}{10}\: , \: R \: = \: 9}

Now , go to back studies we have a statement

If the smallest number which when divided by p,q and r leaves remainder x,y and z where (p-x) = ( q-y) = (r-z) = constant then the number will be LCM(p,q,r) - constant

here ,

2-1 = 1 , 3-2 = 1 , 4-3 = 1 to 10-9 = 1

We get \bold{1} as a constant.

So

\bold{x\: =\: LCM(2,3,4,5,6,7,8,9,10)\:-\:1}

\begin{center}<br />\begin{tabular}{|c|c}<br /><br />2 &amp; 2,3,4,5,6,7,8,9,10\\<br />\cline{1-2}<br />2 &amp; 1,3,2,5,3,7,4,9,5\\<br />\cline{1-2}<br />2 &amp; 1,3,1,5,3,7,2,9,5\\<br />\cline{1-2}<br />3 &amp; 1,3,1,5,3,7,1,9,5\\<br />\cline{1-2}<br />3 &amp; 1,1,1,5,1,7,1,3,5\\<br />\cline{1-2}<br />5 &amp; 1,1,1,5,1,7,1,1,5\\<br />\cline{1-2}<br />7 &amp; 1,1,1,1,1,7,1,1,5\\<br />\cline{1-2}<br />&amp; 1,1,1,1,1,1,1,1,1\\<br />\end{tabular}<br />\end{center}

x = (2×2×2×3×3×5×7)-1

x = 2,520 -1

x = 2,519

The number is \bold{2,519}

\bold{\large{Thanks}}

MOSFET01: i need a coding help
Hakar: awesome dear
Hakar: thanks a lot
Answered by Anonymous
5
● Hope this answer will help u 
N + 1 is divisible by 2, 3, 4, 5, 6, 7, 8, 9 & 10 so clearly one solution would be 10! (ie 10*9*8*7*6*5*4*3*2 = 3628800 for N +1) but this is not the lowest possible. Follow the logic below... 

N + 1 must be a multiple of 2 

N + 1 must be a multiple of 4 but if it is a multiple of 2 & 4 it is necessarily a multiple of 8 

N + 1 must be a multiple of 5 but if it is a multiple of 2 & 5 it is necessarily a multiple of 10 

N + 1 must be a multiple of 6 but if it is a multiple of 2 & 9 (18) it is necessarily a multiple of 6 

N + 1 must be a multiple of 7 

N + 1 must be a multiple of 9 

Hence 2 x 4 x 5 x 7 x 9 =\color{red}\mathcal{2520}
2520  is a multiple of 2, 3, 4, 5, 6, 7, 8, 9 & 10 
 
\color{green}\huge\bold\star\underline\mathcal{neymarjr3}\star
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