Difference between revisions of "2022 AMC 12B Problems/Problem 15"
(Created page with "Problem: One of the following numbers is not divisible by any prime number less than 10. Which is it? <math>\text{(A)} 2^{606}-1 \text{(B)} 2^{606}+1 \text{(C)} 2^{607}-1 \tex...") |
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<cmath>2^7+1=129</cmath> | <cmath>2^7+1=129</cmath> | ||
− | We see that the odd powers of <math>2</math> added with 1 are multiples of three. If we continue this pattern, <math>2^{607}+1</math> will be divisible by <math>3</math>. | + | We see that the odd powers of <math>2</math> added with 1 are multiples of three. If we continue this pattern, <math>2^{607}+1</math> will be divisible by <math>3</math>. (The reason why this pattern works: When you multiply <math>2 \equiv2\text{mod} 3</math> by <math>2</math>, you obtain <math>4 \equiv1 \text{mod} 3</math>. Multiplying by <math>2</math> again, we get <math>1\cdot2\equiv2 \text{mod} 3</math>. We see that in every cycle of two powers of <math>2</math>, it goes from <math>2 \text{mod}3</math> to <math>1 \text{mod}3</math> and back to <math>2 \text{mod}3</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>. | 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>. |
Revision as of 02:02, 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 . Using the sum of cubes factorization
, we have
. Since
ends with a
, and
,
is a multiple of
, which means it is divisible by
.
Since we have eliminated every option except one, is not divisible by any prime less than
.