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Problem #2 · July 29, 2026
Natural numbers
If
A
B
C
‾
+
M
N
‾
=
F
E
D
P
‾
\overline{ABC} + \overline{MN} = \overline{FEDP}
A
B
C
+
M
N
=
F
E
D
P
, calculate the value of
F
M
+
N
+
A
F
F^{M+N} + A^{F}
F
M
+
N
+
A
F
.
A
8
8
8
B
9
9
9
C
✓
10
10
10
D
11
11
11
WORKED EXAMPLE
Solution
When a two-digit number is added to a three-digit number, the sum is at most
999
+
99
=
1098
999+99=1098
999
+
99
=
1098
, so the number
F
E
D
P
‾
\overline{FEDP}
F
E
D
P
is between
1000
1000
1000
and
1098
1098
1098
, and
F
=
1
F=1
F
=
1
. For the sum to have four digits,
A
B
C
‾
≥
1000
−
99
=
901
\overline{ABC}\ge 1000-99=901
A
B
C
≥
1000
−
99
=
901
must hold, so
A
=
9
A=9
A
=
9
. Now
F
M
+
N
+
A
F
=
1
M
+
N
+
9
1
=
1
+
9
=
10
F^{M+N}+A^{F}=1^{M+N}+9^{1}=1+9=10
F
M
+
N
+
A
F
=
1
M
+
N
+
9
1
=
1
+
9
=
10
.
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