2025 AMC 10A Problems/Problem 17
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
?
Video Solution
Solution 1
The problem statement implies that
divides both
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
Note: If an integer
is congruent to an integer
modulo a positive integer
, denoted by
, this means that
divides the difference
~Tacos_are_yummy_1
~andliu766
minor edit by AD_12
Solution 2
We are given that
and
, from which we get
and
.
Substracting the two gives
, which is also divisible by
.
Since
must be greater than
, the only possible divisors of
are
and
. Checking which ones also divide
and
eliminates
and
, but
.
Since both
and
are divisible by
, so is
. Thus,
works, and its tens digit is
.
~Continuous_Pi
~edited by Zhixing
Solution 3
We get that:
and
.
So we also have that:
and
Notice that these are a multiple of
. Now, we subtract these numbers to get
.
We see that
There is a factor of
and
.
and
are multiples of
as well, so our answer is just 45, or
.
~Aarav22
~reformatting by Alzwang
~minor editing by kfclover
Solution 4
We are given that
and
Subtracting, we get
This means
divides
Also, since
we have
Thus
divides both
and
Using the Euclidean Algorithm:
so
Hence
and the tens digit of
is
~samma
Chinese Video Solution
https://www.bilibili.com/video/BV1nLkQBpEXR/
~metrixgo
Video Solution (In 2 Mins)
https://youtu.be/ax76SAmVuYw?si=1YPIk87CnrevwHRm ~ Pi Academy
Video Solution by SpreadTheMathLove
https://www.youtube.com/watch?v=dAeyV60Hu5c
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 | ||
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