#Numbers-8
If the integers m and n are chosen at random between 1 and 100, then the probability that a number of the form 7^m + 7^n is divisible by 5 is:
(a)1/5
(b)1/4
(c)1/7
(d)1/49
Solution-
(7^m + 7^n) mod 5 = 2^m +2^n
Now, the number to be divisible by 5, it must end in 5 or 0.
2^m + 2^n can never end in 5 as it is an even number (minimum value of m,n is 1), so the number must end in 0 if it has to be divisible by 5.
Cyclicity of 2 is 4 (2,4,8,6),
2+8 or 6+4 are the only cases when 2^m +2^n is divisible by 5.
last digit to be 2 and 8, 2^x must be of the form 2^(k+1) and 2^(k+3) respectively
Similarly, last digit to be 4 and 6 it must be of the form 2^(k+2) and 2^(k+4) respectively
Now in both of these cases we will have 25 pairs each,
Required cases = (25*25 + 25*25)*2 [25*25 +25*25 for both cases and multiplied by 2 as m = 1, n=3 or n=1 or m =3 are both valid cases)
Probability = 25*25*2*2/(100*100) = 1/4 ANS
(a)1/5
(b)1/4
(c)1/7
(d)1/49
Solution-
(7^m + 7^n) mod 5 = 2^m +2^n
Now, the number to be divisible by 5, it must end in 5 or 0.
2^m + 2^n can never end in 5 as it is an even number (minimum value of m,n is 1), so the number must end in 0 if it has to be divisible by 5.
Cyclicity of 2 is 4 (2,4,8,6),
2+8 or 6+4 are the only cases when 2^m +2^n is divisible by 5.
last digit to be 2 and 8, 2^x must be of the form 2^(k+1) and 2^(k+3) respectively
Similarly, last digit to be 4 and 6 it must be of the form 2^(k+2) and 2^(k+4) respectively
Now in both of these cases we will have 25 pairs each,
Required cases = (25*25 + 25*25)*2 [25*25 +25*25 for both cases and multiplied by 2 as m = 1, n=3 or n=1 or m =3 are both valid cases)
Probability = 25*25*2*2/(100*100) = 1/4 ANS
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