Difference between revisions of "2018 AMC 10B Problems/Problem 13"
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==Solution 2== | ==Solution 2== | ||
− | Note that <math>10^{2k}+1</math> for some odd <math>k</math> will suffice <math>\mod {101}</math>. Each <math>2k \in \{2, | + | Note that <math>10^{2k}+1</math> for some odd <math>k</math> will suffice <math>\mod {101}</math>. Each <math>2k \in \{2,6,10,\dots,2018\}</math>, so the answer is <math>\boxed{\textbf{(C) } 505}</math> |
(AOPS12142015) | (AOPS12142015) | ||
Revision as of 22:32, 20 December 2018
Problem
How many of the first numbers in the sequence
are divisible by
?
Solution 1
The number is divisible by 101 if and only if
. We note that
, so the powers of 10 are 4-periodic mod 101. It follows that
if and only if
.
In the given list, , the desired exponents are
, and there are
numbers in that list.
Solution 2
Note that for some odd
will suffice
. Each
, so the answer is
(AOPS12142015)
Solution 3
If we divide each number by , we see a pattern occuring in every 4 numbers.
. We divide
by
to get
with
left over. One divisible number will be in the
left over, so out answer is
.
Solution 4
Note that is divisible by
, and thus
is too. We know that
is divisible and
isn't so let us start from
. We subtract
to get 2. Likewise from
we subtract, but we instead subtract
times
or
to get
. We do it again and multiply the 9's by
to get
. Following the same knowledge, we can use mod
to finish the problem. The sequence will just be subtracting 1, multiplying by 10, then adding 1. Thus the sequence is
. Thus it repeats every four. Consider the sequence after the 1st term and we have 2017 numbers. Divide
by four to get
remainder
. Thus the answer is
plus the 1st term or
.
-googleghosh
See Also
2018 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
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