Question

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

Answer 1

n is a multiple of 7, and 1 short of a multiple of 3 or 4 or 5 or 6
Division by 6 takes care of division by 3, so n+1 = 4 x 5 x 6 x K
(where K will ensure it’s a multiple of 7)
So n = 119*K, and 119 is divisible by 7, so K = 1 and n = 119

Answer 2

HEY DEAR.....!!

Let the number of mangoes be x.

when x ÷ 2 leaves remainder as 1.
when x ÷ 3 leaves remainder as 2.
when x ÷ 4 leaves remainder as 3.
when x ÷ 5 leaves remainder as 4.
when x ÷ 6 leaves remainder as 5.
when x ÷ 7 leaves remainder as 0.
⇒ x is divisible by 7.

The remainder in each case is 1 less than the divisor.
⇒ (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.

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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