Understanding the properties of integers - GMAT Quantitative

The product of a group of nonzero numbers is negative if and only if an odd number of these factors is negative. This occurs in each of these scenarios except for one - all of the numbers (eight) being negative.

A prime number is only divisible by the number 1 and itself. Of the integers between 50 and 60, all of the even integers are also divisible by the number 2 so they are not prime numbers. The integers 51 and 57 are divisible by 3. The integer 55 is divisible by 5. The only integers that are prime are 53 and 59. The sum of these two integers is 112.

We must remember that 0 and 1 are NOT prime numbers, but 2 is.

The four smallest prime numbers are 2, 3, 5, and 7. Then 2 * 3 * 5 * 7 = 210.

Note: There are NO negative prime numbers, so we don't have to look for tiny, negative numbers here.

The sum (or difference) or 2 even integers is even. Similarly, the product (or quotient) of 2 even integers is also even; therefore the answer must be , which can be easily checked by plugging in any two even numbers.

For example, if , which is odd.

The sum of 3 consecutive numbers would be

which simplifies into

Set that equation equal to 36 and solve.

To find total number, multiply the number of each item.

The only prime factors are 3 and 5, therefore, 5 will be the greatest prime factor.

The sum is 8,053,467

Since the numerator is always 1 greater than the denominator, we know that for large enough values of , it's never going to be an integer (one will be even, other odd). In fact, there are only 2 cases where this can be done. The first is dividing 0. Since 0 is divisible by every number, if the numerator is 0, then we will still get an integer. Thus one answer is

The other answer occurs as a special case as well. We can divide any number by 1 evenly, so when the denominator is 1 we get an integer:

In every other case, we will have a non-integer.

The product of a group of nonzero numbers is positive if and only if an even number of these factors is negative. This occurs in each of these scenarios.

The easiest way to explain this is by example.

We can use 16, since

This makes 4 the correct choice.

The conditions of the problem can be rewritten as , or .

, meaning that the sum of the two numbers is a multiple of 7. Each of the given choices is a multiple of 7, so any of them can be .

Any integer that is divisible by 5 must end in a 0 or a 5, so any integer which, when divided by 5, yields remainder 3 must end in 3 or 8. Since the cubes of 3 and 8 are 27 and 512, the cube of an integer that ends in one of these digits must end in 2 or 7, meaning that, when divided by 5, the remainder will be 2.

, or, equivalently, . This makes 936 a four-digit number when written in base six. The first digit is equal to the number of times 216 divides into 936.

,

4 is the first digit of this number.

Divide 735 by 8. The remainder is the last digit of the base-eight representation.

The correct choice is 7.

For a number to be divisible by 6, it must be both even and a multiple of 3. None of the choices can be eliminated by that first criterion. But to fit the second criterion, the digit sum must be divisible by 3. We need only add the nonzero digits of each choice, remembering to include the leading 1 in the sum:

Of these choices, only 56 passes the divisibility test, so it is the correct choice.

One way to do this:

Divide 117 by 8. The remainder will be the last digit.

Now divide the quotient by 8. This remainder will be the second-to-last digit.

The quotient is less than 8, so it will be the first digit. The base-eight equivalent of 117 is

All of these expressions are equivalent to the correct quotient. But a number in scientific notation takes the form

, where

Only takes this form.

If a number has a 5 in its base-six representation, then the number divided by 6 yields a remainder of 5. We test each of these five numbers by dividing each by 6 and noting the remainder:

Among the five choices, only 743 yields remainder 5 when divided by 6, so this is the correct choice.