Manipal Entrance Test 2014 Math Paper

Since, m and n are selected between 1 and 100, hence sample space =100×100Also, 71=1,72=49,73=343, 74=2401 75=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 =

100×25

100×100

=

1

4