2017 AMC 10B Problems/Problem 1
Problem
Mary thought of a positive two-digit number. She multiplied it by and added
. Then she switched the digits of the result, obtaining a number between
and
, inclusive. What was Mary's number?
Solution
Solution 1
Let her -digit number be
. Multiplying by
makes it a multiple of
, meaning that the sum of its digits is divisible by
. Adding on
increases the sum of the digits by
and reversing the digits keeps the sum of the digits the same; this means that the resulting number must be
more than a multiple of
. There are two such numbers between
and
:
and
Now that we have narrowed down the choices, we can simply test the answers to see which one will provide a two-digit number when the steps are reversed:
For
we reverse the digits, resulting in
Subtracting
, we get
We can already see that dividing this by
will not be a two-digit number, so
does not meet our requirements.
Therefore, the answer must be the reversed steps applied to
We have the following:
Therefore, our answer is
.
Solution 2
Working backwards, we reverse the digits of each number from ~
and subtract
from each, so we have
The only numbers from this list that are divisible by
are
and
. We divide both by
, yielding
and
. Since
is not a two-digit number, the answer is
.
Solution 3
You can just plug in the numbers to see which one works. When you get to , you multiply by
and add
to get
. When you reverse the digits of
, you get
, which is within the given range. Thus, the answer is
.
See Also
2017 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Doesn't Exist |
Followed by Problem 2 | |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.