Difference between revisions of "2003 AMC 10B Problems/Problem 25"
m |
m (→Solution 4(extremely fast)) |
||
Line 49: | Line 49: | ||
<math>999 < 300c+123 < 10000</math> so <math>876 < 300c < 9877</math>. Dividing by <math>300</math> gives <math>2.92 < c < 32.9222</math> so <math>3\le c \le 32</math>. This gives <math>32-3+1 = \boxed{\textbf{(B)} \ 30 }</math> | <math>999 < 300c+123 < 10000</math> so <math>876 < 300c < 9877</math>. Dividing by <math>300</math> gives <math>2.92 < c < 32.9222</math> so <math>3\le c \le 32</math>. This gives <math>32-3+1 = \boxed{\textbf{(B)} \ 30 }</math> | ||
− | ===Solution 4(extremely fast)=== | + | ===Solution 4 (extremely fast)=== |
The number is in the form <math>xy23</math> | The number is in the form <math>xy23</math> |
Revision as of 16:49, 1 January 2020
Contents
Problem 25
How many distinct four-digit numbers are divisible by and have as their last two digits?
Solution
Solution 1
To test if a number is divisible by three, you add up the digits of the number. If the sum is divisible by three, then the original number is a multiple of three. If the sum is too large, you can repeat the process until you can tell whether it is a multiple of three. This is the basis for the solution. Since the last two digits are , the sum of the digits is (therefore it is not divisible by three). However certain numbers can be added to make the sum of the digits a multiple of three.
,
, and so on.
However since the largest four-digit number ending with is , the maximum sum is
.
Using that process we can fairly quickly compile a list of the sum of the first two digits of the number.
Now we find all the two-digit numbers that have any of the sums shown above. We can do this by listing all the two digit numbers in separate cases.
And finally, we add the number of elements in each set.
Solution 2
A number divisible by has all its digits add to a multiple of The last two digits are and and add up to Therefore the first two digits must add up to digits (including ) are are and are The following combinations are equivalent to :
Let the first term in each combination represent the thousands digit and the second term represent the hundreds digit. We can use this to find the total amount of four-digit numbers.
Solution 3
We have the following: and . Then for some integers and . Taking mod 3 gives: so for some integer . But so . Bounding this gives us: so . Dividing by gives so . This gives
Solution 4 (extremely fast)
The number is in the form
Notice that the number is divisible by if the sum of the digits of is divisible by
Our first number that is divisible by is , next is .. Notice goes from
Hence, there are distinct four digit numbers.
Solution 5 (Even Faster than extremely fast)
Following the form in Solution 4, notice that to satisfy our condition. Choose a value of from 1 to 9. For , there are exactly 4 values of . For the remaining 6 digits, there are 3 choices for y. So our answer is .
Solution 6 (Even Faster than Even Faster than Extremely Fast)
There are possible values for the first two digits. One-third of them yield a multiple of , so the answer is
See also
2003 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 24 |
Followed by Last problem | |
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.