How to Solve an Integer or Consecutive Integer SAT Math Problem

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Word Problem: Integer or Consecutive Integer
Type I: Identify and solve the integer and consecutive integer math problems that appear on every
SAT test.
Step 1
Know your terms.
Integers: whole numbers with no decimals or fractions (ex. -1, 0, 1)
Positive Integers: whole numbers that are positive (ex. 1, 2, 3)
Negative Integers: whole numbers that are negative (ex. -1, -2, -3)
Zero: an integer that is neither positive nor negative
Consecutive Integers: integers in a row, in units of 1 (ex. 97, 98, 99)
Step 2
Take the terms from the problem so that you can correctly translate the English of the
problem into math.
Example #1:
The sum of 3 consecutive integers is 1,200. What is the value of the largest of these integers?
Solution 1
3 consecutive integers = x, x + 1, x + 2
x + x + 1 + x + 2 = 1,200
3x = 1,197
x = 399
So the largest integer = x + 2 = 399 + 2 = 401
Solution 2
3 consecutive integers = ___, ___, ___
average = 1,200 ÷ 3 = [ ___, 400, ___ ]
largest integer = [ 399, 400, 401 ] = 401
TYPE 2: Identify and solve even/odd integer problems.
Step 1
Know your terms.
Even Integers: an integer that is divisible by two (ex. 6,8,10)
Odd Integers: an integer that is not divisible by two (ex. 7,9,11)
Consecutive Even Integers: even integers that increase by 2 (ex. 2,4,6)
Consecutive Odd Integers: odd integers that increase by 2 (ex. 3,5,7)
Algebra for Consecutive Even/Odd Integers: x, x + 2, x + 4...
Step 2
Know your rules.
Addition
Even + Even = Even
Even + Odd = Odd
Odd + Odd = Even
Subtraction
Even - Even = Even
Even - Odd = Odd
Odd - Odd = Even
Multiplication
Even × Even = Even
Even × Odd = Even
Odd × Odd = Odd
Example #2: EVEN / ODD INTEGERS
If q is an even integer, which of the expressions results in an even integer?
(A) q(12) - 7
(B) 5 + 7q
(C) 5q - 3q
(D) 2q + 3
(E) 4q + 1
Solution 1:Using Even / Odd Rules
(A) even(even) - odd = even - odd = odd
(B) odd + odd(even) = odd + even = odd
(C) odd(even) - odd(even) = even - even = even
(D) even(even) + odd = even + odd = odd
(E) even(even) + odd = even + odd = odd
Solution 2: Using Plug-In (using q = 2)
(A) 2(12) - 7 = 17 = odd
(B) 5 + (7)(2) = 5 + 14 = 19 = odd
(C) (5)(2) - (3)(2) = 10 - 6 = 4 = even
(D) (2)(2) + 3 = 4 + 3 = 7 = odd
(E) (4)(2) + 1 = 8 + 1 = 9 = odd
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