Question 1047141

The number {{{n}}} is such that it has to satisfy these two series
a. {{{5m + 1}}}
b. {{{7j + 3}}}

To solve these types first take the {{{LCM}}} of the multiples {{{5}}} and {{7}}} here = {{{35}}}.

So the resulting series will be:
{{{35i + r}}}

Now we find {{{r}}}. To do this set {{{i=0}}} and try to find the first term in the series {{{a}}} and {{{b}}} that is {{{same}}}.

series a. {{{6}}},{{{11}}},{{{16}}},{{{21}}},{{{26}}},{{{highlight(31)}}}
series b. {{{10}}},{{{17}}},{{{24}}},{{{highlight(31)}}},{{{38}}}


 So it matches at {{{highlight(31)}}}. So {{{r=31}}}

Therefore the resulting series is :
{{{35i + 31}}} 

So the smallest number which needs to be added to this is {{{highlight(4)}}} to make it divisible by {{{35}}}.

Therefore, answer is {{{k=4 }}}