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## MTEL General Curriculum Mathematics Practice

Question 1 |

#### Exactly one of the numbers below is a prime number. Which one is it?

\( \large511 \) Hint: Divisible by 7. | |

\( \large517\) Hint: Divisible by 11. | |

\( \large519\) Hint: Divisible by 3. | |

\( \large521\) |

Question 2 |

#### The letters A, and B represent digits (possibly equal) in the ten digit number x=1,438,152,A3B. For which values of A and B is x divisible by 12, but not by 9?

\( \large A = 0, B = 4\) Hint: Digits add to 31, so not divisible by 3, so not divisible by 12. | |

\( \large A = 7, B = 2\) Hint: Digits add to 36, so divisible by 9. | |

\( \large A = 0, B = 6\) Hint: Digits add to 33, divisible by 3, not 9. Last digits are 36, so divisible by 4, and hence by 12. | |

\( \large A = 4, B = 8\) Hint: Digits add to 39, divisible by 3, not 9. Last digits are 38, so not divisible by 4, so not divisible by 12. |

Question 3 |

#### The letters A, B, and C represent digits (possibly equal) in the twelve digit number x=111,111,111,ABC. For which values of A, B, and C is x divisible by 40?

\( \large A = 3, B = 2, C=0\) Hint: Note that it doesn't matter what the first 9 digits are, since 1000 is divisible by 40, so DEF,GHI,JKL,000 is divisible by 40 - we need to check the last 3. | |

\( \large A = 0, B = 0, C=4\) Hint: Not divisible by 10, since it doesn't end in 0. | |

\( \large A = 4, B = 2, C=0\) Hint: Divisible by 10 and by 4, but not by 40, as it's not divisible by 8. Look at 40 as the product of powers of primes -- 8 x 5, and check each. To check 8, either check whether 420 is divisible by 8, or take ones place + twice tens place + 4 * hundreds place = 18, which is not divisible by 8. | |

\( \large A =1, B=0, C=0\) Hint: Divisible by 10 and by 4, but not by 40, as it's not divisible by 8. Look at 40 as the product of powers of primes -- 8 x 5, and check each. To check 8, either check whether 100 is divisible by 8, or take ones place + twice tens place + 4 * hundreds place = 4, which is not divisible by 8. |

Question 4 |

#### The prime factorization of n can be written as n=pqr, where p, q, and r are distinct prime numbers. How many factors does n have, including 1 and itself?

\( \large3\) Hint: 1, p, q, r, and pqr are already 5, so this isn't enough. You might try plugging in p=2, q=3, and r=5 to help with this problem. | |

\( \large5\) Hint: Don't forget pq, etc. You might try plugging in p=2, q=3, and r=5 to help with this problem. | |

\( \large6\) Hint: You might try plugging in p=2, q=3, and r=5 to help with this problem. | |

\( \large8\) Hint: 1, p, q, r, pq, pr, qr, pqr. |

Question 5 |

#### How many factors does 80 have?

\( \large8\) Hint: Don't forget 1 and 80. | |

\( \large9\) Hint: Only perfect squares have an odd number of factors -- otherwise factors come in pairs. | |

\( \large10\) Hint: 1,2,4,5,8,10,16,20,40,80 | |

\( \large12\) Hint: Did you count a number twice? Include a number that isn't a factor? |

Question 6 |

#### What is the greatest common factor of 540 and 216?

\( \large{{2}^{2}}\cdot {{3}^{3}}\) Hint: One way to solve this is to factor both numbers: \(540=2^2 \cdot 3^3 \cdot 5\) and \(216=2^3 \cdot 3^3\). Then take the smaller power for each prime that is a factor of both numbers. | |

\( \large2\cdot 3\) Hint: This is a common factor of both numbers, but it's not the greatest common factor. | |

\( \large{{2}^{3}}\cdot {{3}^{3}}\) Hint: \(2^3 = 8\) is not a factor of 540. | |

\( \large{{2}^{2}}\cdot {{3}^{2}}\) Hint: This is a common factor of both numbers, but it's not the greatest common factor. |

Question 7 |

#### What is the least common multiple of 540 and 216?

\( \large{{2}^{5}}\cdot {{3}^{6}}\cdot 5\) Hint: This is the product of the numbers, not the LCM. | |

\( \large{{2}^{3}}\cdot {{3}^{3}}\cdot 5\) Hint: One way to solve this is to factor both numbers: \(540=2^2 \cdot 3^3 \cdot 5\) and \(216=2^3 \cdot 3^3\). Then for each prime that's a factor of either number, use the largest exponent that appears in one of the factorizations. You can also take the product of the two numbers divided by their GCD. | |

\( \large{{2}^{2}}\cdot {{3}^{3}}\cdot 5\) Hint: 216 is a multiple of 8. | |

\( \large{{2}^{2}}\cdot {{3}^{2}}\cdot {{5}^{2}}\) Hint: Not a multiple of 216 and not a multiple of 540. |

Question 8 |

#### Elena is going to use a calculator to check whether or not 267 is prime. She will pick certain divisors, and then find 267 divided by each, and see if she gets a whole number. If she never gets a whole number, then she€™s found a prime. Which numbers does Elena NEED to check before she can stop checking and be sure she has a prime?

## All natural numbers from 2 to 266.Hint: She only needs to check primes -- checking the prime factors of any composite is enough to look for divisors. As a test taking strategy, the other three choices involve primes, so worth thinking about. | |

## All primes from 2 to 266 .Hint: Remember, factors come in pairs (except for square root factors), so she would first find the smaller of the pair and wouldn't need to check the larger. | |

## All primes from 2 to 133 .Hint: She doesn't need to check this high. Factors come in pairs, and something over 100 is going to be paired with something less than 3, so she will find that earlier. | |

## All primes from \( \large 2\) to \( \large \sqrt{267}\).Hint: \(\sqrt{267} \times \sqrt{267}=267\). Any other pair of factors will have one factor less than \( \sqrt{267}\) and one greater, so she only needs to check up to \( \sqrt{267}\). |

Question 9 |

#### P is a prime number that divides 240. Which of the following must be true?

## P divides 30Hint: 2, 3, and 5 are the prime factors of 240, and all divide 30. | |

## P divides 48Hint: P=5 doesn't work. | |

## P divides 75Hint: P=2 doesn't work. | |

## P divides 80Hint: P=3 doesn't work. |

Question 10 |

#### M is a multiple of 26. Which of the following cannot be true?

## M is odd.Hint: All multiples of 26 are also multiples of 2, so they must be even. | |

## M is a multiple of 3.Hint: 3 x 26 is a multiple of both 3 and 26. | |

## M is 26.Hint: 1 x 26 is a multiple of 26. | |

## M is 0.Hint: 0 x 26 is a multiple of 26. |

Question 11 |

#### The chairs in a large room can be arranged in rows of 18, 25, or 60 with no chairs left over. If C is the smallest possible number of chairs in the room, which of the following inequalities does C satisfy?

\( \large C\le 300\) Hint: Find the LCM. | |

\( \large 300 < C \le 500 \) Hint: Find the LCM. | |

\( \large 500 < C \le 700 \) Hint: Find the LCM. | |

\( \large C>700\) Hint: The LCM is 900, which is the smallest number of chairs. |

Question 12 |

#### The least common multiple of 60 and N is 1260. Which of the following could be the prime factorization of N?

\( \large2\cdot 5\cdot 7\) Hint: 1260 is divisible by 9 and 60 is not, so N must be divisible by 9 for 1260 to be the LCM. | |

\( \large{{2}^{3}}\cdot {{3}^{2}}\cdot 5 \cdot 7\) Hint: 1260 is not divisible by 8, so it isn't a multiple of this N. | |

\( \large3 \cdot 5 \cdot 7\) Hint: 1260 is divisible by 9 and 60 is not, so N must be divisible by 9 for 1260 to be the LCM. | |

\( \large{{3}^{2}}\cdot 5\cdot 7\) Hint: \(1260=2^2 \cdot 3^2 \cdot 5 \cdot 7\) and \(60=2^2 \cdot 3 \cdot 5\). In order for 1260 to be the LCM, N has to be a multiple of \(3^2\) and of 7 (because 60 is not a multiple of either of these). N also cannot introduce a factor that would require the LCM to be larger (as in choice b). |

Question 13 |

#### A biology class requires a lab fee, which is a whole number of dollars, and the same amount for all students. On Monday the instructor collected $70 in fees, on Tuesday she collected $126, and on Wednesday she collected $266. What is the largest possible amount the fee could be?

## $2Hint: A possible fee, but not the largest possible fee. Check the other choices to see which are factors of all three numbers. | |

## $7Hint: A possible fee, but not the largest possible fee. Check the other choices to see which are factors of all three numbers. | |

## $14Hint: This is the greatest common factor of 70, 126, and 266. | |

## $70Hint: Not a factor of 126 or 266, so couldn't be correct. |

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