how many two digits numbers have remainder 1 when they are divided by 5
Answers
11,16,21,26,.......86,91,96
In this A.P :
a = 11
d = 16-11 = 5
l = 96
n = ?
Tn = a+(n-1)d
96 = 11+(n-1)5
n-1 = 85/5
n = 17+1
n = 18
First you need to know that inorder that a number be dividible by 5,it must have either 0 or 5 at its units place.
Now in this question we need to find 2 digit numbers leaving the remainder 1 after dividing by 5.
Notice,such numbers must have 1 or 6 at their unit's place.
Thus,the number of favourable choices for such numbers ,at the unit's place is 2 (Either 1 or 6 can come at the unit's place)
And for tenth place we can choose any number out of 1,2,3,4,5,6,7,8,or 9.
So,favourable choices for tenth place are 9.
Now,by multiplicative law of probability,we have
Total number of 2 digit numbers that leave the remainder 1 on dividing by 5 are:
2X9=18 numbers.
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