Example 51: The number of ordered pairs ( (m, n), m, n ) ( in{1,2, ldots, 100} ) such that ( 7^{m}+7^{n} ) is divisible by 5 is
(a) 1250
(b) 2000
(c) 2500
(d) 5000

( A n s )
(c)
responding to ( r=1,2,3 ) and 4 respectively.) Thus, ( 7^{m}+7^{n} ) cannot end in 5 for any values of ( m, n in N . ) In other words, for ( 7^{m}+7^{n} ) to be divisible by ( 5, ) it should end in ( 0 . ) For ( 7^{m}+7^{n} ) to end in ( 0, ) the forms of ( m ) and ( n ) should be as follows:

hline & ( m ) & ( n )
hline 1 & ( 4 r ) & ( 4 s+2 )
hline 2 & ( 4 r+1 ) & ( 4 s+3 )
hline 3 & ( 4 r+2 ) & ( 4 s )
hline 4 & ( 4 r+3 ) & ( 4 s+1 )
hline
end{tabular}
Thus, for a given value of ( m ) there are just 25 values of ( n ) for which ( 7^{m}+7^{n} ) ends in ( 0 . ) [For instance, if ( m=4 r, ) then ( n= ) ( 2,6,10, ldots, 98] )
( therefore quad ) there are ( 100 times 25=2500 ) ordered pairs ( (m, n) ) for which ( 7^{m}+7^{n} ) is divisible by 5