Integers - GRE Quantitative Reasoning

First we need to find out how many possible numbers there are. The number of possible four-digit numbers with four different digits is simply 4 * 4 * 4 * 4 = 256.

To find the sum, the formula we must remember is sum = average * number of values. The last piece that's missing in this formula is the average. To find this, we can average the first and last number, since the numbers are consecutive. The smallest number that can be created from 1, 2, 3, and 4 is 1111, and the largest number possible is 4444. Then the average is (1111 + 4444)/2.

So sum = 5555/2 * 256 = 711,040.

The distance between two points is always positive. We calculate lx2 - x1l, which will give us the distance between the points.

|14- (-4)| = |14+4| = |18| = 18

To solve, pick numbers to represent and . Let and . Now try each of the equations given:

.

Only works and is thus our answer.

There are 50 positive, odd numbers less than 100, and 45 of them are 2 digit numbers.

For the sum of 7 consecutive even integers to be zero, the only sequence possible is –6, –4, –2, 0, 2, 4, 6. This can be determined algebraically by assigning the lowest number in the sequence to be “y” and adding 2 for each consecutive even integer, and then setting this equal to zero.

y, y + 2, y + 4, y + 6 . . .

The product of any number and 0 is 0.

Therefore, column B must be greater.

If and are both even whole numbers, then their addition must be an even whole number as well. Although is an even number, it is not a whole number and could therefore not be a solution. This means the only possible solution would be .

Plugging in the values given we arrive at the total fruit John has:

A good note about adding even numbers--any even numbers--is that if you add two even numbers, their sum will ALWAYS be an even number.

To deal with a problem with this many digits, often the best strategy is to line up one number over the other, then add the places one at a time. Don't forget to carry a one every time the addition goes over ten. Also, note that any time you add two even numbers, their sum will ALWAYS be an even number.

For the sum of 7 consecutive even integers to be zero, the only sequence possible is –6, –4, –2, 0, 2, 4, 6. This can be determined algebraically by assigning the lowest number in the sequence to be “y” and adding 2 for each consecutive even integer, and then setting this equal to zero.

y, y + 2, y + 4, y + 6 . . .

The product of any number and 0 is 0.

Therefore, column B must be greater.

If and are both even whole numbers, then their addition must be an even whole number as well. Although is an even number, it is not a whole number and could therefore not be a solution. This means the only possible solution would be .

Plugging in the values given we arrive at the total fruit John has:

A good note about adding even numbers--any even numbers--is that if you add two even numbers, their sum will ALWAYS be an even number.

To deal with a problem with this many digits, often the best strategy is to line up one number over the other, then add the places one at a time. Don't forget to carry a one every time the addition goes over ten. Also, note that any time you add two even numbers, their sum will ALWAYS be an even number.

For the sum of 7 consecutive even integers to be zero, the only sequence possible is –6, –4, –2, 0, 2, 4, 6. This can be determined algebraically by assigning the lowest number in the sequence to be “y” and adding 2 for each consecutive even integer, and then setting this equal to zero.

y, y + 2, y + 4, y + 6 . . .

The product of any number and 0 is 0.

Therefore, column B must be greater.

If and are both even whole numbers, then their addition must be an even whole number as well. Although is an even number, it is not a whole number and could therefore not be a solution. This means the only possible solution would be .

Plugging in the values given we arrive at the total fruit John has:

A good note about adding even numbers--any even numbers--is that if you add two even numbers, their sum will ALWAYS be an even number.

To deal with a problem with this many digits, often the best strategy is to line up one number over the other, then add the places one at a time. Don't forget to carry a one every time the addition goes over ten. Also, note that any time you add two even numbers, their sum will ALWAYS be an even number.

Simply recognize that logically, the participation of either sport is non-exclusive, that is, just because people took track and field does not necessarily mean they did not take swimming as well. As such, those 51 who took swimming could have all potentially done track and field, which means all 51 students.

There are 120 students in the math club and 150 students in the chess club, for a total membership of 270. However, 100 students are in both clubs, which means they are counted twice. You simply subtract 100 from 270, which will give you a total of 170 different students participating in both clubs. This means that the remaining 130 students do not participate in either club.

When adding integers, one needs to pay close attention to the sign. When you add a negative integer, it's the same thing as subtracting that integer. Therefore: