I am a positive integer less than 100.
two more than my value is a multiple of 6.
the sum of my digits is a multiple of 7.
I must show all my work
This kind of problem can't be solved by writing and solving one or two equations. You have to use "trial and error" to see what works.
Let x be the ten's digit and y be the one's digit. The actual number is 10x + y
x + y is either 7 or 14, since it must be divisible by 7 and cannot exceed 18 (which you would get with two nines)
10x + y + 2 is divisible by 6.
Possible x,y combinations for which x+y is divisble by 7 are:
(1,6), (2,5), (3,4), (4,3), (5,2), (6,1), (5,9), (6,8), (7,7), (8,6), (9,5)
Now see which one of those has 10x +y +2 divisible by 6. It looks like the number could be 16 or 34 or 52.
To verify if a number is a multiple of 6, you need to check if it is divisible by both 2 and 3. Let's check each of the possible numbers:
1. For the number 16:
- The sum of the digits is 1 + 6 = 7, which is divisible by 7.
- 16 + 2 = 18, which is divisible by 6. Therefore, 16 is a valid solution.
2. For the number 34:
- The sum of the digits is 3 + 4 = 7, which is divisible by 7.
- 34 + 2 = 36, which is divisible by 6. Therefore, 34 is a valid solution.
3. For the number 52:
- The sum of the digits is 5 + 2 = 7, which is divisible by 7.
- 52 + 2 = 54, which is not divisible by 6. Therefore, 52 is not a valid solution.
Based on these calculations, the possible numbers are 16 and 34.