Arithmetic - GMAT Quantitative

Card 0 of 4488

Question

What is the product of all of the prime numbers between 20 and 30?

Answer

There are two prime numbers between 20 and 30; they are 23 and 29. Their product:

$23 \cdot 29 =667$

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Question

You are given that the product of eight nonzero numbers is negative. Which of the following is not possible?

Answer

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.

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Question

What is the sum of the odd numbers from $-155$ to $160$ , inclusive?

Answer

We can divide this group number into two parts. The first part is $\left { -155, 155\right }$ and the second part is $\left { 156, 160\right }$ .

The first group is symmetrical, so the sum of this group of numbers is 0. Now for the second part there are only two numbers left according to the question (sum of odd numbers), which is 157 and 159.

Therefore, the answer is .

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Question

What is the sum of the prime numbers that are greater than 50 but less than 60?

Answer

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.

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Question

What is the product of the four smallest prime numbers?

Answer

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.

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Question

If a and b are even integers, what must be odd?

Answer

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 $\dpi{100} \small a+b-1$ , which can be easily checked by plugging in any two even numbers.

For example, if $\dpi{100} \small a=2\ and\ b=4,\ a+b-1=2+4-1=5$ , which is odd.

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Question

Three consecutive numbers add up to 36. What is the smallest number?

Answer

The sum of 3 consecutive numbers would be

which simplifies into

Set that equation equal to 36 and solve.

$3n + 3 = 36 \rightarrow 3n = 33 \rightarrow n = 11$

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Question

Becky has to choose from 4 pairs of pants, 6 shirts, and 2 pairs of shoes for an interview. If an outfit consists of 1 pair of pants, 1 pair of shoes, and 1 shirt, how many options does she have?

Answer

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

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Question

What is the greatest prime factor of

Answer

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

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Question

Solve:

$(8\times 10^6)+(5\times 10^4)+(3\times 10^3)+(4\times 10^2)+(6\times 10)+(7\times 10^0)=$

Answer

$8\times 10^6=8,000,000$

$5\times 10^4=50,000$

$3\times 10^3=3,000$

$4\times 10^2=400$

$6\times 10=60$

$7\times 10^0=7$

The sum is 8,053,467

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Question

For how many integers, , is $\frac{x-1}{x-2}$ an integer?

Answer

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 $x=1 \Rightarrow \frac{(1)-1}{(1)-2}=\frac{0}{-1} = 0$

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:

$x=3 \Rightarrow \frac{(3)-1}{(3)-2} = \frac{2}{1} = 2$

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

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Question

You are given that the product of eight numbers, each of which is nonzero, is positive. Which of the following is not possible?

Answer

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.

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Question

A positive integer divided by 9 yields remainder 7. What remainder does $N^{2}$ divided by 9 yield?

Answer

The easiest way to explain this is by example.

We can use 16, since $16 \div 9 = 1 \textrm{ R } 7$

$16 ^{2 } = 256$

$256 \div 9 = 28 \textrm{ R } 4$

This makes 4 the correct choice.

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Question

If a positive integer is divided by another positive integer , then the quotient is 6, and there is no remainder.

Which of these choices is a possible value of ?

Answer

The conditions of the problem can be rewritten as $N \div M = 6$ , 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 .

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Question

A positive integer divided by 5 yields remainder 3. What remainder does $N^{3}$ divided by 5 yield?

Answer

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.

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Question

What is the first digit in the base-six representation of the number 936?

Answer

, or, equivalently, $6^{3}< 936 < 6^{4}$ . 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.

$936 \div 216 = 4 \textrm{ R } 72$ ,

4 is the first digit of this number.

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Question

What is the last digit in the base-eight representation of the number 735?

Answer

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

$735 \div 8 = 91 \textrm{ R } 7$

The correct choice is 7.

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Question

A one-hundred-digit integer that is divisible by 6 begins with a 1; its next ninety-seven digits are all zeroes. Which of the following could be its last two digits?

Answer

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.

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Question

Rewrite 117 as a number in base eight.

Answer

One way to do this:

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

$117 \div 8 = 14 \textrm{ R }5$

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

$14 \div 8 = 1 \textrm{ R }6$

The quotient is less than 8, so it will be the first digit. The base-eight equivalent of 117 is $165_{\textrm{eight}}$

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Question

Express the quotient in scientific notation:

$\frac{ 5 \times 10^{18} }{ 8\times 10^{7} }$

Answer

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

$a \times 10^{n}$ , where $1 \leq |a| < 10$

Only $6.25 \times 10^{10}$ takes this form.

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