Tag: history of mathematics

We have,

Given that:

$587\times 999=587\times (1000-1)$
$=587\times 1000-587\times 1$
$=587000-587$
$=586413$

$93521\times 625=935421\times {5}^{4}=935421\times {(\cfrac{10}{2})}^{4}$
$\cfrac{935421\times{10}^{4}}{{2}^{4}}=\cfrac{9354210000}{16}$
$=584638125$

$2056\times 987=2056\times (1000-13)$
$=2056\times 1000-2056\times 13$
$=2056000-26728$
$=2029272$.

$72519\times 9999=72519\times(10000-1)$
$=72519\times 10000-72519\times1$
$=725190000-72519$
$=725117481$

$12345679\times 72=123456789\times (70+2)$
$=12345679\times 70+123445679\times 2$
$=864197530+24691358$
$=888888888$

By hit and trial , we find that
$47619\times=333333$.

$(112\times {5}^{4})=112\times { \left( \cfrac { 10 }{ 2 }  \right)  }^{ 4 }=\cfrac { 112\times { 10 }^{ 4 } }{ { 2 }^{ 4 } } =\cfrac { 1120000 }{ 16 } =70000$

Required numbers are $10,15,20,25,......96$
This is an A.P. in which $a=10,d=5$ and $l=95$
${t} _{n}=95$ $\Rightarrow$ $a+(n-1)d=95$
$\Rightarrow$ $10+(n-1)\times 5=95$
$\Rightarrow$ $(n-1)\times 5=85$
$\Rightarrow$ $(n-1)=17$
$\Rightarrow$ $n=18$
$\therefore$ Required Sum $=\cfrac { n }{ 2 } (a+l)=\cfrac { 18 }{ 2 } \times (10+95)=(95\times 105)=945$