SAT Math 5.3 Algebra and Functions
|Elementary Number Theory||Factors, Multiples, and Remainders|
|Heart of Algebra||Systems of linear equations in two variables (word problems)|
|Polynomials and Rational Expressions||Remainder Theorem|
|Product Type||SAT Math|
|SAT Math||Algebra and Functions|
All right, so we’ve got some mystery number, and we know the remainder is 1 when you divide it by 11.
Now we want to tack on 10 more and do the same… and we need to determine the new remainder.
We need a quotient… we’ll call that k.
Let’s work in reverse. But be safe – check the rear view mirror first.
Multiplying both sides by 11 and then adding back the remainder gives us n = 11k + 1.
So now we’re looking for the remainder when n + 10 is divided by 11.
Adding 10 to both sides, n + 10 = 11k + 11.
Factoring out the 11, so n + 10 = 11(k + 1).
We want to figure out what the remainder is of (n + 10) divided by 11...
…so dividing both sides of the equation by 11 makes the left side exactly what we’re
looking for. When n + 10 is divided by 11 the quotient
is k + 1 and the remainder is 0. Looks like out answer is (A).
As in, “Auto repair.”