nislam
could anyone please give the detail solution of the problem?
Sure!
So, we have integers from 150 to 300 all inclusive
Given that it should leave a remainder of 4 when divided by 6 and a remainder of 2 when divided by 8
Let us examine both seperately
Remainder of 4 with 6: So, we know that \(150\) is completely divisible by \(6\) so leaves no remainder so we can say that \(154\) leaves a remainder of \(4\) with \(6\)
Now since 154 leaves a remainder of 4 when divided by 6, so if we keep adding 6 it will keep giving us integers which leave a remainder of 4 when divided by 6
\(154, 160, 166, 172, 178, 184, 190, 196, 202...\) and so on
Remainder of 2 with 8: So, we know that 160 is completely divisible by 8 so leaves no remainder so we can say that 152 is also divisible by 8 which means that
Edit: 154
leaves a remainder of \(2\) with \(8\)
Now since 154 leaves a remainder of 2 when divided by 8, so if we keep adding 8 it will keep giving us integers which leave a remainder of 2 when divided by 8
\(154, 162, 170, 178, 186, 194, 202...\) and so on
So if you observe both lists you see that \(154, 178, 202\) are common in the
incomplete list and one thing you notice is that the difference between these subsequent terms is always \(24\)
So we can say that after 154, if we keep adding 24 we will get all the terms who leave a remainder of 4 with 6 and a remainder of 2 with 8
And since there are 151 terms in total, adding 24 six times to 154 will give us the list
\(154, 178, 202, 226, 250, 274, 298\)
\(7\) numbers
Answer - DPosted from my mobile device