The answer are
A. 5
B. 6
C. 0
D. 4
From left to right the first three digits of a 4 digits number add up to 8 which digits could be in the ones place if the 4 digit number is divisible by 6
3 answers
For any number to be divisible by 6, it must be divisible by both 2 and 3
To be divisible by 2, the unit digit must be even or zero, so that eliminates A - 5.
To be divisible by 3, the sum of the digits must be a multiple of 3 (divisible by 3)
Since we have a sum of 8 for the first 3 digits we need a sum of:
9, 12, 15, 18 etc
to get sum of 9, the last digit must be 1, but it can't be odd
to get a sum of 12, the last digit must be 4.
That could be a choice
to get a sum of 15, the last digit must be 7, no good
to get a sum of 18, our last digit must be 10, which is not single digit.
We can stop here.
The last digit has to be 4, which is choice D
To be divisible by 2, the unit digit must be even or zero, so that eliminates A - 5.
To be divisible by 3, the sum of the digits must be a multiple of 3 (divisible by 3)
Since we have a sum of 8 for the first 3 digits we need a sum of:
9, 12, 15, 18 etc
to get sum of 9, the last digit must be 1, but it can't be odd
to get a sum of 12, the last digit must be 4.
That could be a choice
to get a sum of 15, the last digit must be 7, no good
to get a sum of 18, our last digit must be 10, which is not single digit.
We can stop here.
The last digit has to be 4, which is choice D
Reiny can you help me with 5 questions
3 answers