Divisibility Algebra Quiz: Master Divisibility Algebra Quiz

Check 24: digit sum = 2 + 4 = 6, divisible by 3 but not by 9 since 6 is not a multiple of 9. Option A: 18 has digit sum 9, divisible by 9. Option C: 27 has digit sum 9, divisible by 9. Option D: 45 has digit sum 9, divisible by 9. Only 24 satisfies divisibility by 3 but not 9.

The expression n = 9k + 1 is already in the form divisor times quotient plus remainder. The divisor is 9, the quotient is k, and the remainder is 1. Option A gives remainder 0, meaning divisibility by 9. Option C gives remainder 8. Option D gives 9, which is not a valid remainder since remainders must be less than the divisor.

A number ending in 0, 2, 4, 6, or 8 is even and divisible by 2, confirming A. The digit sum test for 3 works because 10 is congruent to 1 mod 3, confirming B. Ending in 0 means the number is a multiple of 10, confirming C. The digit sum test for 9 works because 10 is congruent to 1 mod 9, confirming D. Option E is false — even numbers like 2, 4, 8, 10 are not divisible by 3.

LCM(2, 3) = 6 because 2 and 3 share no common factors other than 1. Numbers divisible by both 2 and 3 must be multiples of their LCM. Writing N = 6k captures all such numbers. Option A gives multiples of 2 only. Option B gives multiples of 3 only. Option C gives multiples of 5, which has no relationship to divisibility by 2 and 3.

Substitute x = 1: 7(1) + 2 = 9. Since 9 = 3 times 3, it is divisible by 3. Option B gives 7(2) + 2 = 16, not divisible by 3. Option C gives 7(3) + 2 = 23, not divisible by 3. Option D gives 7(5) + 2 = 37, not divisible by 3. Only x = 1 produces a multiple of 3.

Substitute x = 2: 5(2) + 2 = 10 + 2 = 12. Since 12 = 3 times 4, it is divisible by 3. Option A gives 5(1) + 2 = 7, not divisible by 3. Option C gives 5(4) + 2 = 22, not divisible by 3. Option D gives 5(6) + 2 = 32, not divisible by 3. Only x = 2 produces a multiple of 3.

The answer is True. Factor out 5: 10k + 5 = 5(2k + 1). Since 5 divides the factored form, every number of this form is divisible by 5. This also shows that numbers of this form are odd multiples of 5 — they are divisible by 5 but not by 10.

LCM(3, 5) = 15 because 3 and 5 share no common factors. Numbers divisible by both must be multiples of their LCM. X = 15n captures exactly all such numbers. Option A gives multiples of 5 only. Option B gives multiples of 3 only. Option D gives multiples of 7, which is unrelated to divisibility by 3 and 5.

Numbers divisible by 4 satisfy N = 4m. Numbers leaving remainder 1 mod 3 satisfy N = 3j + 1. By the Chinese Remainder Theorem, the combined form is N = 12k + 4 since 4 is divisible by 4 and 4 mod 3 = 1. Option A gives remainder 1 mod 3 but no guarantee of divisibility by 4. Option B gives multiples of 4 only. Option C gives remainder 3 mod 4, not divisibility by 4.

The division algorithm gives N = 8k + 3 where k is any integer. The 8 is the divisor and 3 is the remainder. Option A gives exact multiples of 8 with remainder 0. Option B gives remainder 1. Option D reverses the roles of divisor and remainder, giving a form that does not represent division by 8.

Divisible by 4 means n is an integer multiple of 4. Writing N = 4k for some integer k captures exactly all multiples of 4. Option A gives multiples of 2, which includes odd multiples of 2 not divisible by 4. Option B gives multiples of 3. Option D gives multiples of 5. Only N = 4k correctly represents divisibility by 4.

The answer is True. Since 9 = 3 times 3, any multiple of 9 can be written as 9k = 3 times (3k), which is clearly a multiple of 3. Divisibility by a number implies divisibility by all of its factors. The converse is not true — for example 6 is divisible by 3 but not by 9.

Check 42: 42 divided by 6 = 7 exactly, so divisible by 6. 42 divided by 5 = 8 remainder 2, satisfying the second condition. Option A: 10 divided by 6 is not an integer. Option B: 22 divided by 6 is not an integer. Option C: 32 divided by 6 is not an integer. Only 42 satisfies both conditions simultaneously.

Odd numbers are integers that are not divisible by 2. Every odd number is one more than an even number, so they can all be written as 2k + 1 for some integer k. Option A gives even numbers. Option C gives multiples of 3, which includes both even and odd numbers. Option D gives multiples of 5, also a mix of parities.

12 = 3 times 4, and since gcd(3, 4) = 1, divisibility by both 3 and 4 together guarantees divisibility by 12. Option A: LCM(2, 5) = 10, not 12. Option B: LCM(3, 9) = 9, not 12, since 9 is already a multiple of 3. Option D: LCM(4, 5) = 20, not 12. Only 3 and 4 produce LCM 12.

LCM(2, 5) = 10 because 2 and 5 share no common factors. Numbers divisible by both 2 and 5 must be multiples of their LCM. X = 10n captures exactly all such numbers. Option A gives multiples of 2 only. Option B gives multiples of 5 only. Option D gives multiples of 7, which has no relationship to divisibility by 2 and 5.

The division algorithm gives x = 7k + 4 for integer k, where 4 is the remainder. Option A gives exact multiples of 7 with remainder 0. Option B gives remainder 1. Option C gives remainder 3. Only x = 7k + 4 correctly encodes a remainder of 4 when dividing by 7.

Even numbers are integers divisible by 2, which means they can all be written as 2 times some integer n. Option A gives 2n + 1, which produces odd numbers not even ones. Option B gives multiples of 3, which includes both even and odd numbers. Option C gives multiples of 4, which is a subset of even numbers but misses numbers like 2, 6, 10. Only X = 2n captures all even numbers.

The answer is True. Since 6 = 2 times 3, we can write 6k = 2 times (3k), confirming divisibility by 2. We can also write 6k = 3 times (2k), confirming divisibility by 3. Any multiple of 6 is automatically a multiple of both its factors 2 and 3.

The division algorithm states that n = divisor times quotient plus remainder. Dividing by 5 with remainder 2 gives N = 5k + 2. Option A gives exact multiples of 5 with remainder 0. Option B gives remainder 1. Option D subtracts 2 rather than adding, giving remainder 3 when written in standard form since 5k - 2 = 5(k-1) + 3.