Number Theory 693744.There are clearly more 2’s than 5’s in the prime factorization ofn!, so it suffices to
solve the equation
⌊n
5
⌋
+
⌊n
52⌋
+
⌊n
53⌋
+··· = 1000.
On the one hand,
⌊n
5⌋
+
⌊n
52⌋
+
⌊n
53⌋
+···<
n
5+
n
52+
n
53+··· =
n
5·
1
1 −^15
=
n
4,
and hencen>4000. On the other hand, using the inequalitya>a−1, we have
1000 >(n
5− 1
)
+
(n
52− 1
)
+
(n
53− 1
)
+
(n
54− 1
)
+
(n
55− 1
)
=
n
5(
1 +
1
5
+
1
52
+
1
53
+
1
54
)
− 5 =
n
5·
1 −
( 1
5) 5
1 −^15
− 5 ,
so
n<1005 · 4 · 3125
3124
< 4022.
We have narrowed down our search to{ 4001 , 4002 ,..., 4021 }. Checking each case
with Polignac’s formula, we find that the only solutions aren=4005, 4006, 4007, 4008,
and 4009.
745.Polignac’s formula implies that the exponent of the number 2 inn!is
⌊n
2
⌋
+
⌊n
22⌋
+
⌊n
23⌋
+···.
Because
n
2+
n
22+
n
23
+··· =nand not all terms in this infinite sum are integers, it follows thatnis strictly greater than
the exponent of 2 inn!, and the claim is proved.
(Mathematics Competition, Soviet Union, 1971)
746.Letpbe a prime number. The power ofpin lcm( 1 , 2 ,...,ni)is equal tokif and
only if
⌊
n
pk+^1
⌋
<i≤⌊
n
pk⌋
.
Hence the power ofpin the expression on the right-hand side is