Number Properties and Divisibility: Lesson Test

Answer: A

Correct answer: 53. Test small primes.

are composite. No prime up to 7 divides 53, so its only factors are 1 and 53.

• A (correct): Its only factors are 1 and 53, so it has exactly two factors and is prime.
• B: Being odd does not make it prime; 49 = 7 × 7 is composite.
• C: Its digits add to 12, so it is divisible by 3; 57 = 3 × 19 is composite.
• D: Its digits add to 6, so it is divisible by 3; 51 = 3 × 17 is composite.

Answer: B

Correct answer: 2,754. Add the digits:

and 18 is divisible by 9, so 2,754 is divisible by 9. Check:

• A: Its digit sum is 15, which passes the test for 3 but not the test for 9.
• B (correct): Its digit sum is 18, a multiple of 9, and 2,754 ÷ 9 = 306.
• C: Ending in 5 shows divisibility by 5, but the digit sum of 10 is not a multiple of 9.
• D: Its digit sum is 12, which is divisible by 3 but not by 9.

Answer: B

Correct answer: 13. Simplify inside the bars first: |6 − 15| = |−9| = 9 and |−4| = 4. Then

• A: This ignores the bars on the first term, using −9 + 4 instead of 9 + 4.
• B (correct): Inside first gives |−9| = 9, and adding |−4| = 4 gives 13.
• C: This subtracts 4 instead of adding the absolute value |−4| = 4.
• D: This drops all the signs before simplifying, adding 6 + 15 + 4.

Answer: A

Correct answer: 2. The largest multiple of 9 not above 83 is 81 (9 × 9). Then

so the quotient is 9 with remainder 2. Check:

• A (correct): 9 × 9 = 81 and 83 − 81 = 2, which is smaller than the divisor.
• B: This is how far 83 is below 90, the next multiple of 9, not what is left over.
• C: This uses 9 × 8 = 72; a remainder of 11 is larger than the divisor, so another 9 fits.
• D: This is the quotient, not the amount left over.

Answer: A

Correct answer: 48. A multiple of 12 is 12 times a whole number. Since

the number 48 is a multiple of 12. The numbers 4 and 6 are factors of 12, and 66 is not divisible by 12.

• A (correct): 12 × 4 = 48, so 48 is a multiple of 12.
• B: 6 divides 12, so it is a factor of 12, not a multiple of it.
• C: 4 divides 12, so it is a factor of 12, not a multiple of it.
• D: 66 = 6 × 11 is a multiple of 6, but 66 ÷ 12 leaves a remainder of 6.

Answer: D

Correct answer: 24. Factor both:

Take the shared primes at their smaller powers:

• A: 12 divides both numbers, but it is not the greatest common factor; 24 is larger and also works.
• B: This uses only one 2 and one 3 instead of the lower power of each shared prime.
• C: This is the least common multiple, which uses the highest powers of every prime.
• D (correct): Shared primes at the lower powers give $2^3\times 3=24$, and 24 divides both numbers.

Answer: C

Correct answer: 36. Factor both:

Use every prime at its highest power:

• A: Both 12 and 18 divide 72, but 36 is a smaller common multiple.
• B: This is the greatest common factor, not the least common multiple.
• C (correct): $2^2\times 3^2=36$, the smallest number both 12 and 18 divide into.
• D: This is the product 12 × 18, a common multiple but not the smallest one.

Answer: D

Correct answer: 7,416. A number is divisible by 12 when it passes the tests for both 3 and 4. For 7,416, the last two digits form 16, a multiple of 4, and the digit sum is

a multiple of 3. Check:

• A: 24 is a multiple of 4, but the digit sum 13 is not divisible by 3.
• B: Being even with digit sum 12 makes it divisible by 6, but 50 is not a multiple of 4.
• C: The digit sum 18 passes the test for 3, but 38 is not a multiple of 4.
• D (correct): 16 is a multiple of 4 and the digit sum 18 is a multiple of 3, so it passes both tests.

Answer: B

Correct answer: 7,392. Alternately subtract and add the digits from the left. For 7,392:

a multiple of 11, so 7,392 is divisible by 11. Check:

• A: 4 − 5 + 7 − 3 = 3, which is not 0 or a multiple of 11.
• B (correct): 7 − 3 + 9 − 2 = 11, a multiple of 11, and 7,392 ÷ 11 = 672.
• C: 5 − 2 + 8 − 1 = 10, which is not 0 or a multiple of 11.
• D: 9 − 1 + 4 − 6 = 6, which is not 0 or a multiple of 11.

Answer: C

Correct answer: 2² × 3² × 5. Split

Grouping equal primes gives the result.

• A: This product is only 60; one factor of 3 is missing.
• B: This product is 120, not 180; it has an extra 2 and is missing a 3.
• C (correct): Every factor is prime and the product is 4 × 9 × 5 = 180.
• D: The product is 180, but 10 is not prime, so this is not a prime factorization.

Answer: C

Correct answer: 29. The sum of three consecutive integers is three times the middle one, so the middle integer is

The integers are 27, 28 and 29, and the largest is 29.

• A: This is the middle integer; the question asks for the largest.
• B: Starting from 28 and counting two more steps overshoots; 28, 29 and 30 sum to 87.
• C (correct): The middle integer is 84 ÷ 3 = 28, so the three are 27, 28 and 29.
• D: This is the smallest of the three integers, not the largest.

Answer: D

Correct answer: 3. The greatest number of identical kits is the GCF of 84 and 126. Since

the GCF is

kits. Each kit gets

snack bars.

• A: This uses 21 kits, a common factor but not the greatest one, so fewer kits than possible are made.
• B: This is the number of water bottles per kit (84 ÷ 42), not snack bars.
• C: This is the number of kits, not the number of snack bars in each kit.
• D (correct): With 42 kits, each kit holds 126 ÷ 42 = 3 snack bars.

Answer: B

Correct answer: 120 seconds. Use the LCM. With

take each prime at its highest power:

• A: This multiplies 6 × 8 × 10; it is a common multiple but counts the shared factors of 2 too many times.
• B (correct): $2^3\times 3\times 5=120$ is the smallest number that 6, 8 and 10 all divide.
• C: This multiplies the LCM of 6 and 8, which is 24, by 10 without removing the shared factor of 2.
• D: This is the greatest common factor; a repeating schedule calls for the LCM.

Answer: A

Correct answer: a + b. Odd plus even is odd, so a + b is always odd. The product ab has an even factor, so it is even. 3a is odd, and odd plus 1 is even.

is a product of neighboring integers, so it is even.

• A (correct): An odd number plus an even number is always odd.
• B: A product with at least one even factor is even, and b is even.
• C: 3a is odd because both factors are odd, and odd plus 1 gives an even number.
• D: This equals a(a + 1), a product of two neighboring integers, which is always even.

Answer: C

Correct answer: 1. Write

Since 21k is a multiple of 7, only 15 matters, and

remainder 1. For example, n = 12 gives

• A: This is 7 − 5, how far n is from the next multiple of 7, not the remainder of 3n.
• B: Tripling the remainder gives 15, but a remainder must be smaller than the divisor, so it must be reduced.
• C (correct): 3n = 21k + 15, and 15 leaves remainder 1 when divided by 7.
• D: This assumes the remainder stays the same when n is multiplied by 3.

Answer: D

Correct answer: 4,326. Check the last digit for 2 and the digit sum for 3. Only 4,326 is even with a digit sum (

) divisible by 3, and

• A: It is even, but its digit sum is 10, which is not a multiple of 3.
• B: Its digit sum of 12 passes the test for 3, but it ends in 1, so it is odd and fails the test for 2.
• C: It ends in 5, so it is odd, and its digit sum of 13 is not a multiple of 3.
• D (correct): It ends in 6, so it is even, and its digits add to 15, a multiple of 3, so it passes both tests.

Answer: A

Correct answer: 8. A factor divides the number with no remainder. Since

8 is a factor of 56.

• A (correct): 56 ÷ 8 = 7 with no remainder, so 8 divides 56 exactly.
• B: 56 ÷ 12 is 4 with remainder 8, so 12 does not divide 56.
• C: 112 is a multiple of 56 (56 × 2), not a factor; a factor cannot be larger than the number.
• D: 56 ÷ 6 is 9 with remainder 2, so 6 is not a factor.

Answer: D

Correct answer: 21 pieces. Equal pieces as long as possible call for the GCF. 96 = $2^5$ × 3 and 156 = $2^2$ × 3 × 13, so the GCF is $2^2$ × 3 = 12 inches.

Then

for

pieces.

• A: 12 is the length of each piece in inches, not the number of pieces.
• B: This uses 6-inch pieces; 6 is a common factor, but not the greatest one.
• C: This counts only the pieces cut from the 156-inch pipe and leaves out the 8 from the 96-inch pipe.
• D (correct): The longest equal length is the GCF, 12 inches, giving 8 + 13 = 21 pieces.

Answer: B

Correct answer: 9:12 a.m. Repeating schedules call for the LCM. 18 = 2 × $3^2$ and 24 = $2^3$ × 3, so the LCM is $2^3$ × $3^2$ = 72 minutes.

72 minutes is 1 hour 12 minutes, so the buses next leave together at 9:12 a.m.

• A: This adds 18 and 24 to get 42 minutes, but the buses do not line up after 42 minutes.
• B (correct): The LCM of 18 and 24 is 72 minutes, and 72 minutes after 8:00 is 9:12.
• C: This uses the product 18 × 24 = 432 minutes, a common multiple but not the least one.
• D: This uses the GCF, 6 minutes, which is the tool for equal groups, not repeating schedules.

Answer: D

Correct answer: 2² × 3² × 7. Split and keep splitting until every factor is prime:

• A: The product is 252, but 21 = 3 × 7 is not prime, so the factorization is not finished.
• B: This has one 2 too many; it equals 504, not 252.
• C: This is missing a factor of 2; it equals 126, half of 252.
• D (correct): 4 × 9 × 7 = 252, and 2, 3 and 7 are all prime.

Answer: B

Correct answer: 51. A composite number has more than two factors. The digits of 51 add to 6, so 3 divides it:

The other choices have no factors besides 1 and themselves.

• A: 29 has only the factors 1 and 29, so it is prime.
• B (correct): 51 = 3 × 17, so it has factors other than 1 and itself.
• C: 31 has only the factors 1 and 31, so it is prime.
• D: 37 has only the factors 1 and 37, so it is prime.

Answer: A

Correct answer: 5. The largest multiple of 8 not above 125 is

The remainder is

Check:

• A (correct): 8 × 15 = 120 and 125 − 120 = 5, which is less than 8.
• B: 15 is the quotient, not the remainder.
• C: This rounds the quotient up to 16 and finds 128 − 125 = 3; the multiple of 8 must not exceed 125.
• D: This is the decimal part of 125 ÷ 8 = 15.625, not the whole-number amount left over.

Answer: C

Correct answer: 44. For three equally spaced integers the sum is three times the middle one, so the middle is

Consecutive even integers are 2 apart, so they are 44, 46 and 48, and

The smallest is 44.

• A: 46 is the middle integer, not the smallest.
• B: This treats them as ordinary consecutive integers (45, 46, 47), but the question asks for even integers.
• C (correct): The middle integer is 138 ÷ 3 = 46, so the even integers are 44, 46 and 48.
• D: 48 is the largest of the three integers.

Answer: C

Correct answer: 2,124. Test 4 with the last two digits and 8 with the last three digits. For 2,124, 24 is a multiple of 4, but 124 leaves a remainder of 4 when divided by 8, so 2,124 is divisible by 4 only.

• A: It is divisible by 4, but 416 = 8 × 52, so it is also divisible by 8.
• B: Its last two digits, 34, are not a multiple of 4.
• C (correct): Its last two digits, 24, are a multiple of 4, but its last three digits, 124, are not a multiple of 8 (124 ÷ 8 = 15.5).
• D: Its last two digits, 30, are not a multiple of 4.

Answer: B

Correct answer: x + y is even. A product is odd only if every factor is odd, so x and y are both odd. Odd plus odd is even, so x + y must be even.

• A: Both numbers must be odd, and the sum of two odd numbers is even, not odd.
• B (correct): An odd product means both x and y are odd, and odd plus odd is even.
• C: If x were even, the product xy would be even; x must be odd.
• D: xy is odd, so adding 1 gives an even number.

Answer: C

Correct answer: 18. Factor into primes: 54 = 2 × $3^3$ and 90 = 2 × $3^2$ × 5. Multiply the shared primes at their lower powers: 2 × $3^2$ = 18.

• A: 9 divides both numbers, but it misses the shared factor of 2.
• B: 6 is a common factor, but it uses only one 3 when both numbers share two.
• C (correct): 54 = 2 × $3^3$ and 90 = 2 × $3^2$ × 5 share 2 × $3^2$ = 18.
• D: 270 is the least common multiple, not the greatest common factor.

Answer: A

Correct answer: 4,730. A number divisible by 2 ends in an even digit, and one divisible by 5 ends in 0 or 5. Only a last digit of 0 does both, so 4,730 is the answer.

• A (correct): It ends in 0, which passes the tests for 2, 5 and 10.
• B: It ends in 5, so it is divisible by 5 but is odd.
• C: It ends in 2, so it is even but not divisible by 5.
• D: It ends in 4, so it is even but not divisible by 5.

Answer: C

Correct answer: 56. Use every prime at its highest power: 8 = $2^3$ and

so the LCM is $2^3$ × 7 = 56. Both 8 × 7 and 14 × 4 equal 56.

• A: 112 = 8 × 14 is a common multiple, but not the smallest one.
• B: 2 is the greatest common factor, not the least common multiple.
• C (correct): 8 = $2^3$ and 14 = 2 × 7, so the LCM is $2^3$ × 7 = 56.
• D: 28 is a multiple of 14 but not of 8 (28 ÷ 8 = 3.5).

Answer: D

Correct answer: 4,059. Apply both tests to each number. For 4,059 the digit sum is

divisible by 3, and the alternating sum

means it is divisible by 11.

Check:

• A: It passes the test for 11 (5 − 3 + 0 − 2 = 0), but its digit sum of 10 is not a multiple of 3.
• B: Its digit sum of 12 passes the test for 3, but 4 − 2 + 5 − 1 = 6 fails the test for 11.
• C: It passes the test for 11 (6 − 1 + 8 − 2 = 11), but its digit sum of 17 is not a multiple of 3.
• D (correct): Its digit sum is 18, a multiple of 3, and 4 − 0 + 5 − 9 = 0, so it passes the test for 11.

Answer: B

Correct answer: 3. Simplify inside each set of bars first:

Take absolute values to get 9 and 6, then subtract:

• A: This adds 9 and 6 instead of subtracting the second absolute value.
• B (correct): |3 − 12| = |−9| = 9 and |−2 − 4| = |−6| = 6, so 9 − 6 = 3.
• C: This drops the bars on the first part, using −9 − (−6) = −3, but the absolute value of −9 is 9.
• D: This drops the signs before simplifying, using 3 + 12 = 15 and 2 + 4 = 6.

Answer: A

Correct answer: Friday. Days of the week repeat every 7 days, so only the remainder matters.

with remainder 2 (

).

Two days after Wednesday is Friday.

• A (correct): 100 = 7 × 14 + 2, so after 14 full weeks there are 2 extra days, and Wednesday plus 2 days is Friday.
• B: This would be right only if 100 were a multiple of 7, but 100 ÷ 7 leaves a remainder of 2.
• C: This moves ahead only 1 day; the remainder of 100 ÷ 7 is 2.
• D: This moves ahead 3 days, as if the remainder were 3 instead of 2.

Answer: A

Correct answer: 12. Test divisors from 1 up to about the square root of 72 (about 8.5): 1 × 72, 2 × 36, 3 × 24, 4 × 18, 6 × 12, 8 × 9.

Six pairs of different numbers give 12 factors.

• A (correct): The factor pairs are 1 × 72, 2 × 36, 3 × 24, 4 × 18, 6 × 12 and 8 × 9, giving 12 factors.
• B: 6 is the number of factor pairs; each pair holds two different factors.
• C: This misses a pair, such as 8 and 9, which are found just before stopping at the square root.
• D: This misses one factor, often 1 or 72 itself, which must both be counted.

Answer: C

Correct answer: 2 is the only even prime number. A prime is a whole number greater than 1 with exactly two factors. Any even number above 2 is divisible by 1, 2 and itself, so it is composite. That leaves 2 as the only even prime.

• A: 1 has only one factor, so it is neither prime nor composite; the smallest prime is 2.
• B: Odd numbers such as 9, 15 and 21 are composite.
• C (correct): Every other even number has 2 as a factor, so it has more than two factors.
• D: A prime has exactly two factors, 1 and itself.

Answer: B

Correct answer: 14 groups. Identical groups with nothing left over call for the GCF.

so the GCF is

Each group gets

red and

blue cones.

• A: 7 divides both numbers, but it is not the greatest common factor.
• B (correct): The GCF of 42 and 70 is 14, so she can make 14 groups of 3 red and 5 blue cones.
• C: 8 is the number of cones in each group (3 red + 5 blue), not the number of groups.
• D: 210 is the LCM; it cannot be the number of groups because there are only 112 cones.

Answer: A

Correct answer: 3. The known digits add to

The next multiple of 9 is 18, so

(The one after that, 27, would need k = 12, which is not a single digit.)

• A (correct): The digit sum becomes 3 + 7 + 3 + 5 = 18, a multiple of 9, and 3,735 ÷ 9 = 415.
• B: The digit sum would be 21, which is divisible by 3 but not by 9.
• C: The digit sum would be 15, which is divisible by 3 but not by 9.
• D: The digit sum would be 19, which is not divisible by 9.

Answer: C

Correct answer: 15. A common factor must divide both numbers with no remainder.

so 15 works.

• A: 9 is a factor of 45, but 30 ÷ 9 leaves a remainder of 3.
• B: 6 is a factor of 30, but 45 ÷ 6 leaves a remainder of 3.
• C (correct): 45 = 15 × 3 and 30 = 15 × 2, so 15 divides both exactly.
• D: 90 is a common multiple of 45 and 30, not a common factor.

Answer: B

Correct answer: n(n + 3). Adding an odd number (3) switches parity, so exactly one of n and n + 3 is even. A product with at least one even factor is even. For example, n = 5 gives

• A: n and n + 2 have the same parity; if n is odd, both factors are odd and the product is odd.
• B (correct): n and n + 3 always have opposite parity, so one factor is even and the product is even.
• C: 2n is even, so adding 1 always makes the result odd.
• D: Adding 4 keeps the parity of n, so the result is odd whenever n is odd.

Answer: D

Correct answer: 17. Write the integers as n, n + 1, n + 2 and n + 3. Their sum is

so 4n = 68 and n = 17.

Check:

• A: 18 is the second integer in the list.
• B: 18.5 is the average of the four integers, which lies between the two middle numbers, not an integer in the list.
• C: 20 is the largest of the four integers.
• D (correct): With n + (n + 1) + (n + 2) + (n + 3) = 4n + 6 = 74, n = 17, and 17 + 18 + 19 + 20 = 74.

Answer: D

Correct answer: 7,203. Add the digits: if the sum is a multiple of 3, the number is too. Only 7,203 has a digit sum (12) divisible by 3.

• A: Its digit sum is 16, which is not a multiple of 3.
• B: Its digit sum is 7, which is not a multiple of 3.
• C: Its digit sum is 7, which is not a multiple of 3.
• D (correct): Its digits add to 7 + 2 + 0 + 3 = 12, a multiple of 3, and 7,203 ÷ 3 = 2,401.