GreenlightTestPrep wrote:
In the correctly-performed addition above, each letter represents a different non-zero digit. What is the value of C?
A) 1
B) 2
C) 3
D) 4
E) 5
So this one is very tricky. The explanation is long so bear with me.
Since \(C\) can be anything from 1-9, let's see if we can take a look at the \(A\)'s and \(B\)'s.
Note that since we're dealing with single digits, and they're different, the addition in the second column can't be greater than 20, so the only numbers than can carry over to the first column are 0 and 1. The same goes for the third column: \(C + B + A < 30\), so the numbers that can carry over to the second column are either 0, 1, or 2.
Also note: since A changes to B in the sum of the first column, a number is carried over from the second column's addition. This eliminates the possibility of 0 (no number carried over), which indicates that 1 is the number carried over from the second column's addition.
All this together means two things:
\(A + B >= 10\)
and
\(A + 1 = B\)
With these two in mind, we are narrowed down to three different pairs for \(A\) and \(B\):
5,6
6,7
8,9
Remember these for later, we'll need theseUsing those two equations above and the pairs, we can deduce that:
\(C + B + A = B\)
\(B + A + x = A\)
\(A + 1 = B\)
Where \(x\) in the equation can be 1 or 2 carried over from \(C + B + A = B\).
Note that in these equations the sums B, A, and B in the first, second and third equations respecitively represent the units digit of the sum, not the sum itself.Manipulating the third equation:
\(A = B - 1\)
Now we can substitute in \(B - 1\) wherever \(A\) is:
\(C + B + B - 1 = B\)
\(B + B - 1 + x = B - 1\)
Simplifying:
\(C + B = 1\)
[1]\(B + x = 0\)
[2]Equation [1] is saying that when we add \(C + B\) it will produce a number with a units digit of 1.
Since \(C\) and \(B\) are single digit numbers from 1-9, the only two digit number that can produce a 1 in the units digit given the pairs of numbers highlighted in blue above is
11.
In other words:
\(C + B = 11\)
This means that we can eliminate the pair {5,6}, since, for example: if \(B\) is 5, then \(A\) is 6, but \(C\) must be 6 as well, however we can't have two of the same digits.
This leaves us with pairs {6,7} and {8,9}.
Now lets take a look at
equation [2]:
\(B + x = 0\)
\(x\) is the number that carries over from the third column into the second column. If \(B\) is either 6 or 7, then x must be 4 or 3 respectively, but we know that the number that carries over from the third column's addition must be less than 3
(\(C + B + A < 30\)).
This leaves us with our final pair: {8,9}.
We'll need to bring in the equation from the first column to help narrow down our choice for \(B\):
\(C + B + A = B\)
\(C + B = 11\)
So if \(B = 8\), then \(A = 9\) and \(C = 3\), leading to:
\(3 + 8 + 9 = 20\), so \(B = 0\).
This can't be true, so let's try \(B = 9\):
So if \(B = 9\), then \(A = 8\) and \(C = 2\), leading to:
\(2 + 9 + 8 = 19\), so \(B = 9\), and we carry the 1.
That works!
Continuing on, we note that since \(B = 9\), then \(x = 1\) from:
\(B + x = 0\)
So in the second column:
\(9 + 8 + 1 = 18\), so \(A = 8\), and we carry the 1.
Continuing to the third and last column:
\(A + 1 = B\)
\(8 + 1 = 9\), and \(B = 9\).
And we did it!
C = 2What this looks like when we're done:
892
_89
+_8
989
:D