Difference between revisions of "2022 AMC 12B Problems/Problem 15"
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Next, we examine option B. We see that <math>2^{606}</math> has a units of digits of <math>4</math> (Taking the units digit of the first few powers of two gives a pattern of <math>2, 4, 8, 6, 2, 4, 8, 6, 2, 4, 8, 6,\cdots</math>). Adding <math>1</math> to <math>4</math>, we get <math>5</math>. Since <math>2^{606}+1</math> has a units digit of <math>5</math>, it is divisible by <math>5</math>. | Next, we examine option B. We see that <math>2^{606}</math> has a units of digits of <math>4</math> (Taking the units digit of the first few powers of two gives a pattern of <math>2, 4, 8, 6, 2, 4, 8, 6, 2, 4, 8, 6,\cdots</math>). Adding <math>1</math> to <math>4</math>, we get <math>5</math>. Since <math>2^{606}+1</math> has a units digit of <math>5</math>, it is divisible by <math>5</math>. | ||
− | Lastly, we examine | + | Lastly, we examine option A. Using the difference of cubes factorization <math>a^3-b^3=(a-b)(a^2+ab+b^2)</math>, we have <math>2^{606}-1^3=(2^{202}-1)(2^{404}+2^{202}+1)</math>. Since <math>2^{404}+2^{202}+1\equiv0\text{mod}3</math> (Every term in the sequence is equivalent to <math>1\text{mod}3</math>), <math>2^{606-1}</math> is divisible by <math>3</math>. |
− | Since we have eliminated every option except | + | Since we have eliminated every option except C, <math>\boxed{\text{(C)}2^{607}-1}</math> is not divisible by any prime less than <math>10</math>. |
Revision as of 02:16, 18 November 2022
Problem: One of the following numbers is not divisible by any prime number less than 10. Which is it?
Solution 1 (Process of Elimination)
We examine option E first. has a units digit of
(Taking the units digit of the first few powers of two gives a pattern of
) and
has a units digit of
(Taking the units digit of the first few powers of three gives a pattern of
). Adding
and
together, we get
, which is a multiple of
, meaning that
is divisible by 5.
Next, we examine option D. We take the first few powers of added with
:
We see that the odd powers of added with 1 are multiples of three. If we continue this pattern,
will be divisible by
. (The reason why this pattern works: When you multiply
by
, you obtain
. Multiplying by
again, we get
. We see that in every cycle of two powers of
, it goes from
to
and back to
.)
Next, we examine option B. We see that has a units of digits of
(Taking the units digit of the first few powers of two gives a pattern of
). Adding
to
, we get
. Since
has a units digit of
, it is divisible by
.
Lastly, we examine option A. Using the difference of cubes factorization , we have
. Since
(Every term in the sequence is equivalent to
),
is divisible by
.
Since we have eliminated every option except C, is not divisible by any prime less than
.