ProbabilityHard
Question
If the integers m and n are chosen at random between 1 and 100, then the probability that a number of the form 7m + 7n is divisoble by 5, equals
Options
A.
B.
C.
D.
Solution
71 = 7,72 = 49,73 = 343,74 = 2401, .........
Therefore, for 7r, r∈ N the number ends at unit place 7, 9, 3, 1, 7,.....
∴ 7m + 7n will be divisible by 5 if it end at 5 or 0.
Buit it cannot end at 5.
Also for end at 0.
For this m and n should be as follows
for any given value of m, there will be 25 values of n. Hence, the probabilitty of the required event is
Note:- Power of prime numbers have cyclic numbers in their unit place.
Therefore, for 7r, r∈ N the number ends at unit place 7, 9, 3, 1, 7,.....
∴ 7m + 7n will be divisible by 5 if it end at 5 or 0.
Buit it cannot end at 5.
Also for end at 0.
For this m and n should be as follows
for any given value of m, there will be 25 values of n. Hence, the probabilitty of the required event is
Note:- Power of prime numbers have cyclic numbers in their unit place.
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