2016 AMC 8 Problems/Problem 5
The number is a two-digit number.
• When is divided by , the remainder is .
• When is divided by , the remainder is .
What is the remainder when is divided by ?
Solution 1
From the second bullet point, we know that the second digit must be . Because there is a remainder of when it is divided by , the multiple of must end in a . We now look for this one:
The number satisfies both conditions. We subtract the biggest multiple of less than to get the remainder. Thus, .
Solution 2
We can use modular arithmetic to solve this. Firstly, we can form the equations:
Therefore, and . Since the divisibility rule for is that the last digit has to be , we can say that a number that has a remainder of when divided by ends in .
As the number is more than a multiple of , the multiple of ends in . The numbers that are greater than and end in are: and so on. We can see that is the smallest positive multiple of that ends in a , so must equal .
Now, we need to find the remainder when is divided by . The largest multiple of that is less than is , so is the remainder when is divided by .
Our answer is .
2016 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
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