Since, m and n are selected between 1 and 100, hence sample space =100×100Also, 71=1,72=49,73=343,74=240175=16807 etc. Hence, 1,3,7 and 9 will be the last digits in the power of 7 . Hence, for favourable cases.
‌
‌
nm
‌
‌
↓
1,2
1,2
1,3
...
1,100
2,1
2,2
2,3
...
2,100
100,1
100,2
100,3
...
100,100
for m=1,‌‌n=3,7,11,......97 ∴ favourable cases =25 for m=2,;‌n=4,8,12,10 ∴ favourable cases =25 Similarly for every m. favourable n are 25 . ∴ Total favourable cases =100×25 Hence, required probability =