1988 AIME Problems/Problem 9
Contents
Problem
Find the smallest positive integer whose cube ends in .
Solution 1
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 must be even. . Therefore, , where is any non-negative integer. The number has form . So the minimum .
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
Solution 4 (Bash)
Let this integer be Note that We wish to find the residue of mod Note that using our congruence in mod The residue that works must also satisfy from our original congruence. Noting that (and bashing out the other residues perhaps but they're not that hard), we find that Thus, The residue that works must also satisfy from our original congruence. It is easy to memorize that Also, Finally, as desired. Thus, must satisfy ~samrocksnature
solution 5
This number is in the form of , after binomial expansion, we only want . We realize that are both multiples of , we only need that , so we write
Then, we write so must be a multiple of at least, so after checking, when . So , smallest , the number is
~bluesoul
Solution 6 (A bit of brute force using basic knowledge.)
We do know the unit digit has to be 2, So lets consider the number of the form of .
On cubing , we get a number of the form where the unit digit of must be 8, therefore x can be 4 or 9. But for this value of the hundreds digit won't be 8.
Thus our number must be of the form / . Repeating the above process we get values of as 4 and 1 respectively.
Therefore the smallest number is 192.~ Dubey619
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 | ||
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