Difference between revisions of "1988 AIME Problems/Problem 9"
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Therefore, <math>n = 2 + 5(3+ 5(2+5a_2)) = 125a_2 + 67</math>. <math>n</math> must also be a multiple of <math>8</math>, so <math>125a_2 + 67 \equiv 5a_2 + 3 \pmod 8 \implies a_2 = 1,9,17 \ldots</math>. Therefore, the minimum of <math>n</math> is <math>125 + 67 = \boxed{192}</math>. | Therefore, <math>n = 2 + 5(3+ 5(2+5a_2)) = 125a_2 + 67</math>. <math>n</math> must also be a multiple of <math>8</math>, so <math>125a_2 + 67 \equiv 5a_2 + 3 \pmod 8 \implies a_2 = 1,9,17 \ldots</math>. Therefore, the minimum of <math>n</math> is <math>125 + 67 = \boxed{192}</math>. | ||
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+ | == Solution 3 == | ||
+ | Let <math>x^3 = 1000a + 888</math>. We factor an <math>8</math> out of the right hand side, and we note that <math>x</math> must be of the form <math>x = 2y</math>, where <math>y</math> is a positive integer. Then, this becomes <math>y^3 = 125a + 111</math>. Taking mod <math>5</math>, <math>25</math>, and <math>125</math>, we get <math>y^3 \equiv 1\pmod 5</math>, <math>y^3 \equiv 11\pmod{25}</math>, and <math>y^3 \equiv 111\pmod{125}</math>. | ||
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+ | We can work our way up, and find that <math>y\equiv 1\pmod 5</math>, <math>y\equiv 21\pmod{25}</math>, and finally <math>y\equiv 96\pmod{125}</math>. This gives us our smallest value, <math>y = 96</math>, so <math>x = \boxed{192}</math>, as desired. - Spacesam | ||
== See also == | == See also == |
Revision as of 18:02, 31 July 2019
Problem
Find the smallest positive integer whose cube ends in .
Solution
A little bit of checking tells us that the units digit must be 2. Now our cube must be in the form of ; using the binomial theorem gives us . Since we are looking for the tens digit, we get . This is true if the tens digit is either or . Casework:
- : Then our cube must be in the form of . Hence the lowest possible value for the hundreds digit is , and so is a valid solution.
- : Then our cube is . The lowest possible value for the hundreds digit is , and we get . Hence, since , the answer is
Solution 2
and . due to the last digit of . Let . By expanding, .
By looking at the last digit again, we see , so we let where . Plugging this in to gives . Obviously, , so we let where can be any non-negative integer.
Therefore, . must also be a multiple of , so . Therefore, the minimum of is .
Solution 3
Let . We factor an out of the right hand side, and we note that must be of the form , where is a positive integer. Then, this becomes . Taking mod , , and , we get , , and .
We can work our way up, and find that , , and finally . This gives us our smallest value, , so , as desired. - Spacesam
See also
1988 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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