From a basket of mangoes when counted in twos there was one extra, counted in threes there were two extra, counted in fours there were three extra, counted in fives there were four extra,counted in sixes there were five extra.But counted in sevens there were no extra.Atleast how many mangoes were there in the basket?

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Solution

Let the number of mangoes be $x$ .

when $x \div 2$ leaves remainder as $1$ .
when $x \div 3$ leaves remainder as $2$ .
when $x \div 4$ leaves remainder as $3$ .
when $x \div 5$ leaves remainder as $4$ .
when $x \div 6$ leaves remainder as $5$ .
when $x \div 7$ leaves remainder as $0$ .
$\Rightarrow x$ is divisible by $7$ .
The remainder in each case is $1$ less than the divisor.
$\Rightarrow (x + 1)$ is the LCM of $2, 3, 4, 5$ and $6$ .
LCM of $2, 3, 4, 5$ and $6 = 60$ .
If $x + 1 = 60$ , then $x = 59$ .
But $59$ is not divisible by $7$ .
Move to $60 \times 2 = 120$
If $x + 1 = 120$ , then $x = 119$ .
$119$ is divisible by $7$ also it satisfies all the conditions.
Hence the number of mangoes $= 119$ .

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