How many even 3-digit numbers that is divisible by 10 can be
- Thread starter brucejin
- Start date
mmm4444bot
Super Moderator
- Joined
- Oct 6, 2005
- Messages
- 10,902
I'm thinking that 100 is too many.
I'm also not sure whether or not any of the four digits can be repeated.
If repeats are not allowed, then I think of the senario this way.
There is a box containing four digits.
If a Whole number is divisible by 10, then it's rightmost digit must be zero.
Therefore, one of the four digits in the box must be zero.
So, to start forming three-digit numbers, we pick out the zero and place it in the third position.
This leaves three digits remaining in the box from which to choose for the first position.
After we pick out one of these and place it in the first position, there are two digits remaining in the box from which to choose for the middle position.
Do you think that this explanation matches the given senario?
Hello, brucejin!
Is that the exact wording of the problem?
If so, I have two issues with it.
#1: If a number is divisible by 10, it is automatically even.
. . .The word "even" is unnecessary and possibly confusing.
#2: Among the given four digits, there must be a zero.
. . .Why didn't they simply give us the four digits? .Say, {0, 2, 5, 8}
. . .(This is assuming that they meant 4 different digits.)
Since the 3-digit number is divisible by 10, it must end in zero: ._ _ 0
. . For the 1st digit, there are 3 choices.
. . For the 2nd digit, there are 2 choices.
Therefore, there are: .\(\displaystyle 3\cdot2 \:=\:6\) such numbers.
Is that the exact wording of the problem?
If so, I have two issues with it.
#1: If a number is divisible by 10, it is automatically even.
. . .The word "even" is unnecessary and possibly confusing.
#2: Among the given four digits, there must be a zero.
. . .Why didn't they simply give us the four digits? .Say, {0, 2, 5, 8}
. . .(This is assuming that they meant 4 different digits.)
Since the 3-digit number is divisible by 10, it must end in zero: ._ _ 0
. . For the 1st digit, there are 3 choices.
. . For the 2nd digit, there are 2 choices.
Therefore, there are: .\(\displaystyle 3\cdot2 \:=\:6\) such numbers.
mmm4444bot
Super Moderator
- Joined
- Oct 6, 2005
- Messages
- 10,902
brucejin said:
Okay, this question is clear.
Three-digit numbers sometimes contain repeated digits.
EG: 440, 100, 990
Follow the same logic that Soroban and I discussed.
You understand that the third digit must be zero.
How many choices are there for the first digit?
How many choices are there for the second digit?
(After correcting the posted exercise, your previous answer of 100 is still too high.)