Find a three-digit number divisible by 9 whose first and the last digits are both 1.
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Answer:
To find: 3 digit number divisible by 8!
Lowest digit number = 87×2
=
174
highest 3 digit number = 87×11=
957
applying arithmetic series
l=a+(n−1)d [where, l = last no a = fist no d = difference between number]
957=174+(n−1)×87
n=10
Therefore, there are 10; 3 digit numbers divisible by 87
thanks
Similar questions
Lowest digit number = 87×2
=
174
highest 3 digit number = 87×11=
957
applying arithmetic series
l=a+(n−1)d [where, l = last no a = fist no d = difference between number]
957=174+(n−1)×87
n=10
Therefore, there are 10; 3 digit numbers divisible by 87