2025 AMC 10A Problems/Problem 17: Difference between revisions
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~Continuous_Pi | ~Continuous_Pi | ||
==Solution 3== (Insanely Fast) | |||
We get that: | |||
273436 ≡ 16 (mod N) | |||
272760 ≡ 15 (mod N) | |||
So we also have that: | |||
273420 ≡ 0 (mod N) | |||
272745 ≡ 0 (mod N) | |||
Notice that these are a multiple of 5. Now, we subtract these numbers to get 675 | |||
675 = 25 * 27. We see a factor of 9. | |||
273,420 and 272745 are multiples of 9 so our answer is just <imath>\boxed{\text{(E) }4}5</imath> | |||
~Aarav22 | |||
== Video Solution (In 2 Mins) == | == Video Solution (In 2 Mins) == | ||
Revision as of 19:20, 6 November 2025
(Problem goes here)
Problem
Let
be the unique positive integer such that dividing
by
leaves a remainder of
and dividing
by
leaves a remainder of
. What is the tens digit of
?
Solution 1
The problem statement implies
and
We want to find
that satisfies both of these conditions. Hence, we can just find the greatest common divisor of the two numbers.
by the Euclidean Algorithm, so the answer is
~Tacos_are_yummy_1
Solution 2
We have: 273436 ≡ 16 (mod N) 272760 ≡ 15 (mod N)
First we substract (273436 − 272760) = 676 ≡ 1 (mod N)
So N divides 675. Since N is greater than 16, possible divisors are all greater than 16 and are 25, 27, 45, 75, 135, 225, 675. We then check which ones work. If 273436 ≡ 16 (mod N), then 273420 must be divisible by N. 273420 ÷ 45 = 6076, so N = 45 works. So N = 45, and the tens digit is
.
~Continuous_Pi
==Solution 3== (Insanely Fast)
We get that:
273436 ≡ 16 (mod N)
272760 ≡ 15 (mod N)
So we also have that:
273420 ≡ 0 (mod N)
272745 ≡ 0 (mod N)
Notice that these are a multiple of 5. Now, we subtract these numbers to get 675
675 = 25 * 27. We see a factor of 9.
273,420 and 272745 are multiples of 9 so our answer is just
~Aarav22
Video Solution (In 2 Mins)
https://youtu.be/ax76SAmVuYw?si=1YPIk87CnrevwHRm ~ Pi Academy
Video Solution
~MK
See Also
| 2025 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 16 |
Followed by Problem 18 | |
| 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 | ||
These problems are copyrighted © by the Mathematical Association of America.