Question

# If the integers ‘m’ and ‘n’ are chosen at random from 1 to 100, then the probability that a number of the form of ${{7}^{m}}+{{7}^{n}}$ is divisible by 5, equal to(a) $\dfrac{1}{4}$ (b) $\dfrac{1}{2}$ (c) $\dfrac{1}{8}$ (d) $\dfrac{1}{3}$

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Hint: First write in the form 1(mod 4), 2(mod 4), 3(mod 4), 0(mod 4) then find unit digit of ${{7}^{m}}$ and ${{7}^{n}}$ respectively. Then make a pair of those whose sum gives unit digit 0 or 5. Then find the total number of possible events and the number of favourable events and the probability.

We know that m and n can attain values from 1-100. Then, each ${{7}^{m}}$,${{7}^{n}}$ will have 100 values.
So, total number of values that ${{7}^{m}}+{{7}^{n}}$ can attain is ${{7}^{n}}$ values, that is, ${{7}^{m}}+{{7}^{n}}$ can attain $100\times 100=10000$ values.
So, the total number of events is 10000.
Now we will write numbers wherever possible in terms of a= b (mod c) which relates that if ‘a’ is divided by ‘c’ and ‘b’ will be remainder, for example: 10=2(mod 4), 27=0(mod 3).
Now we will represent ‘n’ in those forms such that there are four different types of ‘n’ or for different ways to represent it.
So, n=1(mod 4)
n=2(mod 4)
n=3(mod 4)
n=0(mod 4)
Now we can say that,
If n=1(mod 4) then the unit digit of ${{7}^{n}}$ will be 7.
If n=2(mod 4) then the unit digit of ${{7}^{n}}$ will be 9.
If n=3(mod 4) then the unit digit of ${{7}^{n}}$ will be 3.
If n=4(mod 4) then the unit digit of ${{7}^{n}}$ will be 1.
Now we can say that there are 25 cases each, for all the four cases with unit digits 7, 9, 1, 3.
So, we know that a number with unit digit as 0 or 5 will be divisible by 5.
The combination of the sum of digits will be (7,7), (7,9),(7,1),(7,3),(9,1),(9,3),(9,9),(3,3),(3,1),(1,1)
So, out of all the pairs only (7,3), (9,1) are the two pairs whose will be divided by 5.
As we know that each number consists of 25 numbers with unit digit (7,3) and (9,1) each, so the total possibility of these cases are $4\times 25\times 25$.
Here two is multiplied because a pair can be represented as (7,3),(3,7) and (9,1),(1,9) will be taken as account.
We know the number of favourable events as$4\times 25\times 25$ and total number of events are $100\times 100$.
So, the probability of finding a number in form of (${{7}^{m}}+{{7}^{n}}$) that be divisible by $5=\dfrac{4\times 25\times 25}{100\times 100}=\dfrac{1}{4}$
Hence, the correct answer is option (a).

Note: Be careful about writing or how to write a=b(mod c)as the significance of this expression is that it represents a number ‘a’ when divided by ‘c’ will leave a remainder ‘b’. While choosing the pairs which satisfy the equation we should be careful that we don’t leave the pair unchecked. For example, a student can take $2\times 25\times 25$, considering only the terms (7,3), (9,1) and forgetting out the terms (3, 7), (1,9).