Difference between revisions of "2022 AMC 10B Problems/Problem 13"
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Let the two primes be <math>a</math> and <math>b</math>. We would have <math>a-b=2</math> and <math>a^{3}-b^{3}=31106</math>. Using difference of cubes, we would have <math>(a-b)(a^{2}+ab+b^{2})=31106</math>. Since we know <math>a-b</math> is equal to <math>2</math>, <math>(a-b)(a^{2}+ab+b^{2})</math> would become <math>2(a^{2}+ab+b^{2})=31106</math>. Simplifying more, we would get <math>a^{2}+ab+b^{2}=15553</math>. | Let the two primes be <math>a</math> and <math>b</math>. We would have <math>a-b=2</math> and <math>a^{3}-b^{3}=31106</math>. Using difference of cubes, we would have <math>(a-b)(a^{2}+ab+b^{2})=31106</math>. Since we know <math>a-b</math> is equal to <math>2</math>, <math>(a-b)(a^{2}+ab+b^{2})</math> would become <math>2(a^{2}+ab+b^{2})=31106</math>. Simplifying more, we would get <math>a^{2}+ab+b^{2}=15553</math>. | ||
− | |||
Now let's introduce another variable. Instead of using <math>a</math> and <math>b</math>, we can express the primes as <math>x+2</math> and <math>x</math> where <math>a</math> is <math>x+2</math> and b is <math>x</math>. Plugging <math>x</math> and <math>x+2</math> in, we would have <math>(x+2)^{2}+x(x+2)+x^{2}</math>. When we expand the parenthesis, it would become <math>x^{2}+4x+4+x^{2}+2x+x^{2}</math>. Then we combine like terms to get <math>3x^{2}+6x+4</math> which equals <math>15553</math>. Then we subtract 4 from both sides to get <math>3x^{2}+6x=15549</math>. Since all three numbers are divisible by 3, we can divide by 3 to get <math>x^{2}+2x=5183</math>. | Now let's introduce another variable. Instead of using <math>a</math> and <math>b</math>, we can express the primes as <math>x+2</math> and <math>x</math> where <math>a</math> is <math>x+2</math> and b is <math>x</math>. Plugging <math>x</math> and <math>x+2</math> in, we would have <math>(x+2)^{2}+x(x+2)+x^{2}</math>. When we expand the parenthesis, it would become <math>x^{2}+4x+4+x^{2}+2x+x^{2}</math>. Then we combine like terms to get <math>3x^{2}+6x+4</math> which equals <math>15553</math>. Then we subtract 4 from both sides to get <math>3x^{2}+6x=15549</math>. Since all three numbers are divisible by 3, we can divide by 3 to get <math>x^{2}+2x=5183</math>. | ||
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~Trex226 | ~Trex226 | ||
+ | |||
+ | A note: If you aren't entirely familiar with square trinomials, after we have <math>x^{2}+2x=5183</math>, we can simply guess and check from here. We know that <math>70^{2} = 4900</math>, and we notice that for <math>x^{2}+2x=5183</math>, <math>x</math> must be a odd number. Thus we can guess that <math>x</math>=71, which proves to be right. Continuing, we then know the larger prime is 73, and we know that 75 and 77 aren't primes, so thus our next bigger prime is 79, and 7+9=16. Thus the answer is <math>\boxed{\textbf{(E) }16}</math>. | ||
+ | |||
+ | ~Rhx | ||
==Solution 2== | ==Solution 2== | ||
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ab = 5183. | ab = 5183. | ||
</cmath> | </cmath> | ||
− | Because we have <math>b | + | Because we have <math>a = b+2</math>, <math>ab = (b+1)^2 - (1)^2</math>. Thus, <math>(b+1)^2 = 5183 + 1 = 5184</math>, so <math>b+1 = 72</math>. This implies <math>a = 73</math>, <math>b = 71</math>, and thus the next biggest prime is <math>79</math>, so our answer is <math>7 + 9 = \boxed{\textbf{(E) }16}</math> |
~mathboy100 | ~mathboy100 | ||
+ | ~ minor edits by AnubhavDey | ||
==Solution 3 (Estimation)== | ==Solution 3 (Estimation)== | ||
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==Solution 4== | ==Solution 4== | ||
− | Let the two primes be <math>x + 1</math> and <math>x - 1</math>. Then, plugging it into the second condition, we get <math>(x + 1)^3 - (x - 1)^3 = 31106.</math> Expanding the left side, <cmath>6x^2 + 2 = 31106 \implies x^2 = 5184.</cmath> Taking the square root of both sides, we get that <math>x = 72</math> and the larger prime is <math>73</math>. The smallest prime larger than <math>73</math> is <math>79</math>, which has a digit sum of <math>7 + | + | Let the two primes be <math>x + 1</math> and <math>x - 1</math>. Then, plugging it into the second condition, we get <math>(x + 1)^3 - (x - 1)^3 = 31106.</math> Expanding the left side, <cmath>6x^2 + 2 = 31106 \implies x^2 = 5184.</cmath> Taking the square root of both sides, we get that <math>x = 72</math> and the larger prime is <math>73</math>. The smallest prime larger than <math>73</math> is <math>79</math>, which has a digit sum of <math>7 + 9 = \boxed{16}.</math> |
- NL008 | - NL008 | ||
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==Video Solution by Interstigation== | ==Video Solution by Interstigation== | ||
https://youtu.be/yLFiSmLJJ5A | https://youtu.be/yLFiSmLJJ5A | ||
+ | |||
+ | ==Video Solution by TheBeautyofMath== | ||
+ | https://youtu.be/Mi2AxPhnRno?t=626 | ||
+ | |||
+ | ~IceMatrix | ||
== See Also == | == See Also == | ||
{{AMC10 box|year=2022|ab=B|num-b=12|num-a=14}} | {{AMC10 box|year=2022|ab=B|num-b=12|num-a=14}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 09:40, 25 July 2024
Contents
Problem
The positive difference between a pair of primes is equal to , and the positive difference between the cubes of the two primes is
. What is the sum of the digits of the least prime that is greater than those two primes?
Solution 1
Let the two primes be and
. We would have
and
. Using difference of cubes, we would have
. Since we know
is equal to
,
would become
. Simplifying more, we would get
.
Now let's introduce another variable. Instead of using and
, we can express the primes as
and
where
is
and b is
. Plugging
and
in, we would have
. When we expand the parenthesis, it would become
. Then we combine like terms to get
which equals
. Then we subtract 4 from both sides to get
. Since all three numbers are divisible by 3, we can divide by 3 to get
.
Notice how if we add 1 to both sides, the left side would become a perfect square trinomial: which is
. Since
is too small to be a valid number, the two primes must be odd, therefore
is the number in the middle of them. Conveniently enough,
so the two numbers are
and
. The next prime number is
, and
so the answer is
.
~Trex226
A note: If you aren't entirely familiar with square trinomials, after we have , we can simply guess and check from here. We know that
, and we notice that for
,
must be a odd number. Thus we can guess that
=71, which proves to be right. Continuing, we then know the larger prime is 73, and we know that 75 and 77 aren't primes, so thus our next bigger prime is 79, and 7+9=16. Thus the answer is
.
~Rhx
Solution 2
Let the two primes be and
, with
being the larger prime. We have
, and
. Using difference of cubes, we obtain
. Now, we use the equation
to obtain
. Hence,
Because we have
,
. Thus,
, so
. This implies
,
, and thus the next biggest prime is
, so our answer is
~mathboy100 ~ minor edits by AnubhavDey
Solution 3 (Estimation)
Let the two primes be and
such that
and
By the difference of cubes formula,
Plugging in and
,
Through the givens, we can see that .
Thus,
Recall that and
. It follows that our primes must be only marginally larger than
, where we conveniently find
The least prime greater than these two primes is
~BrandonZhang202415 ~SwordOfJustice (small edits)
Solution 4
Let the two primes be and
. Then, plugging it into the second condition, we get
Expanding the left side,
Taking the square root of both sides, we get that
and the larger prime is
. The smallest prime larger than
is
, which has a digit sum of
- NL008
Video Solution (⚡️Lightning Fast⚡️)
~Education, the Study of Everything
Video Solution by Interstigation
Video Solution by TheBeautyofMath
https://youtu.be/Mi2AxPhnRno?t=626
~IceMatrix
See Also
2022 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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