Difference between revisions of "2014 AIME I Problems/Problem 8"
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However we dont have to deal with this whole expression but only with its last 4 digits so it is suffices to consider only: | However we dont have to deal with this whole expression but only with its last 4 digits so it is suffices to consider only: | ||
<math>2000ad+2000bc+100c^2+200bd+20cd+d^2</math> | <math>2000ad+2000bc+100c^2+200bd+20cd+d^2</math> | ||
− | + | now we need to compare each decimal digit with <math>1000a+100b+10c+d</math> and see whether the digits are congrount in base 10. | |
we first consider the ones digits: | we first consider the ones digits: | ||
Revision as of 17:20, 8 February 2015
Contents
[hide]Problem 8
The positive integers and both end in the same sequence of four digits when written in base 10, where digit a is not zero. Find the three-digit number .
Solution (general)
We have that
Thus, must be divisible by both and . Note, however, that if either or has both a 5 and a 2 in its factorization, the other must end in either 1 or 9, which is impossible for a number that is divisible by either 2 or 5. Thus, one of them is divisible by , and the other is divisible by . Noting that , we see that 625 would work for , except the thousands digit is 0. The other possibility is that is a multiple of 16 and is a multiple of 625. In order for this to happen, must be congruent to -1 (mod 16). Since , we know that . Thus, , so , and our answer is .
Solution (bashing)
let for positive integer values t,a,b,c,d when we square N we get that
However we dont have to deal with this whole expression but only with its last 4 digits so it is suffices to consider only: now we need to compare each decimal digit with and see whether the digits are congrount in base 10. we first consider the ones digits:
this can happen for only 3 values : 1, 5 and 6
we can try to solve each case
- Case 1
considering the tenths place we have that:
so
considering the hundreds place we have that
so again
now considering the thousands place we have that
so we get but cannot be equal to 0 so we consider
- Case 2
considering the tenths place we have that:
( the extra 20 is carried from which is equal to 25) so
considering the hundreds place we have that
( the extra 100c is carried from the tenths place) so
now considering the thousands place we have that
( the extra 1000b is carried from the hundreds place) so a is equal 0 again
- Case 3
considering the tenths place we have that:
( the extra 20 is carried from which is equal to 25) if then we have
so
considering the hundreds place we have that
( the extra 100c+100 is carried from the tenths place)
if then we have
so
now considering the thousands place we have that
( the extra 1000b+6000 is carried from the hundreds place)
if then we have
so
so we have that the last 4 digits of N are and is equal to
Solution (not bashing)
By the Chinese Remainder Theorem, the equation is equivalent to the two equations: Since and are coprime, the only solutions are when .
Let , . The statement of the Chinese Remainder theorem is that is an isomorphism between the two rings. In this language, the solutions are , , , and . Now we easily see that and . Noting that , it follows that . To compute , note that in , so since is linear in its arguments (by virtue of being an isomorphism), .
The four candidate digit strings are then . Of those, only has nonzero first digit, and therefore the answer is .
See also
2014 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 7 |
Followed by Problem 9 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.