Solution:
For n = 2, n= 4 and n = 5 the values that A assumes are 12+22,12+22+32+42,12+22+32+42+57respectively. Each of these is divisible by 5.
For n = 1 or 3, A takes values 12 and 12+22+32 respectively both of which are not divisible by 5.
So in the set of the 1st 5 natural numbers, 3 numbersare divisible by 5.
For n = 6, 7, 8, 9,10 A behaves in exactly the same manner as for n = 1,2, 3, 4, 5 respectively.
This pattern repeats for the next set of 5 natural numbers and so on.
So for n = 1 to n = 100, A is divisible by 5, in three-fifths of cases. So for 60 values of n A would be divisible by 5.
Since n < 100 and for n = 100, A is divisible by 5, the total number of values that satisfy the condition would be 59.
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