Question 463816: The number 48 is divisible by its unit’s digit, 8. How many two digit numbers are divisible by their units’ digit?
a) 31 b) 41 c) 50 d) 55 e) none of the above
Answer by richard1234(7193)   (Show Source):
You can put this solution on YOUR website! We go case by case, by units digit:
0 - Nothing is "divisible" by 0 since division by zero is undefined.
1 - Any number is divisible by 1, so all of the numbers {11, 21, ..., 91} (9 numbers)
2 - Any number ending in 2 is divisible by 2, {12, 22, ..., 92} (9 numbers)
We continue to go case by case, up to 9:
3 - {33, 63, 93} (3 numbers)
4 - {24, 44, 64, 84} (4 numbers)
5 - {15, 25, ..., 95} (9 numbers)
6 - {36, 66, 96} (3 numbers)
7 - {77} (1 number)
8 - {48, 88} (2 numbers)
9 - {99} (1 numbers)
The total number of numbers is 9+9+3+4+9+3+1+2+1 = 41, answer choice b).
An easy way to generate all numbers is to start with the original number and add any multiple of the LCM of that number and 10. Here, if we use 4, we start with 4 and add multiples of LCM(4, 10). The LCM of 4 and 10 is 20, so we can add any multiple of 20 to get {24, 44, ...}. This would be particularly useful if we wanted to count all the numbers from 1 to 1000 that satisfied this property.
|
|
|
