SOLUTION: find the an odd three digits
number, which is add upto
12, and the diffrnce
between last two digits is
equals to 1st two digits,
what is that num ?
Algebra.Com
Question 830652: find the an odd three digits
number, which is add upto
12, and the diffrnce
between last two digits is
equals to 1st two digits,
what is that num ?
Answer by JulietG(1812) (Show Source): You can put this solution on YOUR website!
Note for future posts: Please take the time to write out the question fully. Tutors are giving you their time, and it helps when we have full information.
.
.
I'm going to make some assumptions, based on what you wrote.
Let a, b, and c be the digits
a + b + c = 12
c - b = a + b
c is an odd number
Given those values, the easiest way is to plug in possibles.
.
The possible digits for c, knowing that the total is odd:
1 (we can toss this one because c-b=a+b. Not possible)
3 (if c is 3, then a + b must be 9. There's no way to make the c-b equation work)
5 (if c is 5, then a + b must be 7. There's no way to make the c-b equation work)
7
9 (we can toss this one because it would necessitate a & b being 1 & 2, which doesn't fit the c-b=a+b)
.
That leaves us 7. If c is 7, then a+b must equal 5. Combinations that would work are 5,0; 4,1; 3,2
We know that c-b = a+5. We've already established that a+b=5. Therefore c-b=5. If C is 7, then B *must* be 2. If B is 2, then a *must* be 3.
.
The solution is 327. <<--
.
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