‘n’ is selected from the set \[\left\{ {1,2,3,...,100} \right\}\]and the number \[{2^n} + {3^n} + {5^n}\]is formed. What are the total number of ways of selecting ‘n’ so that the formed number is divisible by \[4\].
A \[50\]
B \[49\]
C \[48\]
D None of these.

Hint: lection of items from a given collection irrespective of order of section is called combination. Multiples of \[4\]is divisible by \[4\].

Complete step by step solution: The given set is \[\left\{ {1,2,3,...,100} \right\}\]. The format of the number is \[{2^n} + {3^n} + {5^n}\] which is divisible by \[4\].
To find the total number of ways of selecting ‘n’ so that the formed number is divisible by \[4\].
First check \[{2^n}\], It is divisible by \[4\]if \[n > 1\]. For \[n = 1\]the remainder is \[2\].
If \[{3^n}\] is divided by \[4\], it gives remainder \[3\]if n is odd and it gives remainder \[1\] if n is even.
Also, for \[{5^n}\] is divided by \[4\], it gives remainder 1 for any value of n.
Now, the number \[{2^n} + {3^n} + {5^n}\]is divisible by \[4\]when sum of remainders when divided by 4 of \[{2^n},{3^n}\] and \[{5^n}\] is multiple of \[4\].
Find the sum of remainders when divided by 4 of \[{2^n},{3^n}\] and \[{5^n}\] when n is even.
\[0 + 1 + 1 = 2\]
It is not divisible by \[4\]. So, n cannot be even number from given set.
When \[n = 1\] the sum of remainders when divided by 4 of \[{2^n},{3^n}\] and \[{5^n}\]is as follows,
\[2 + 3 + 1 = 6\]
It is not multiple of \[4\]. So, n cannot be \[1\].
Find the sum of remainders when divided by 4 of \[{2^n},{3^n}\] and \[{5^n}\] when n is odd and greater than \[1\].
\[0 + 3 + 1 = 4\]
It is multiple of \[4\]and divisible by \[4\].
In the given set there are \[50\] odd numbers. But when \[n = 1\] ,\[{2^n} + {3^n} + {5^n}\] is not divisible by \[4\].
So, n will be any odd number except \[1\].
The total number of ways of selecting ‘n’ so that the number \[{2^n} + {3^n} + {5^n}\] is divisible by \[4\] will be \[49\].

Option ‘B’ is correct

Note: The common mistake happen here is students consider as \[4\]is even, n can be even which is not true.