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Problem
Problem
Let
∑
k
=
1
10
f
(
a
+
k
)
=
16
(
2
10
−
1
)
\sum_{k=1}^{10} f(a+k) = 16(2^{10} - 1)
∑
k
=
1
10
f
(
a
+
k
)
=
16
(
2
10
−
1
)
, where the function
f
f
f
satisfies
f
(
x
+
y
)
=
f
(
x
)
f
(
y
)
f(x+y) = f(x)f(y)
f
(
x
+
y
)
=
f
(
x
)
f
(
y
)
for all natural numbers
x
,
y
x, y
x
,
y
and
f
(
1
)
=
2
f(1) = 2
f
(
1
)
=
2
. Then the natural number 'a' is: (a) 2 (b) 16 (c) 4 (d) 3
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