2016 AMC 8 Problems/Problem 5

Problem

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 , for a number divisible by ends in zero. Since there is a remainder of when is divided by , the multiple of must end in a for it to have the desired remainder We now look for this one:

The number satisfies both conditions. We subtract the biggest multiple of less than to get the remainder. Thus, .

~CHECKMATE2021

Solution 2

We know that the number has to be one more than a multiple of , because of the remainder of one, and the number has to be more than a multiple of , which means that it has to end in a . Now, if we just list the first few multiples of adding one to the number we get: . As we can see from these numbers, the only one that has a three in the units place is , thus we divide by , getting , hence, . -fn106068

We could also remember that, for a two-digit number to be divisible by , the sum of its digits has to be equal to . Since the number is one more than a multiple of , the multiple we are looking for has a ones digit of , and therefore a tens digit of , and then we could proceed as above. -vaisri

Video Solution

https://youtu.be/d-bCEDoZEjg?si=VFLhpgyJ_vHhE7h3

A solution so simple a 12-year-old made it!

~Elijahman~

Video Solution by OmegaLearn

https://youtu.be/7an5wU9Q5hk?t=574

Video Solution

https://youtu.be/aKWQl7kEMy0

~savannahsolver

Solution 3 We can set up a system of modular congruences

n = 1 (mod 9)
n = 3 (mod 10)

9x + 1 = 3 (mod 10)

   -1   -1

9x = 2(mod 10)

    +70

9x = 72(mod 10) /9 /9

x = 8 (mod 10)

10f + 8 9(10f+8)+1= n 90f + 72 + 1 = n 90f + 73 = n n = 73 (mod 90)

73/11 = 6 r7