Divisibility Shortcuts Quiz: Divisibility Shortcuts for Small Primes

11 is odd, passing the 2 test. Digit sum 1+1=2, not a multiple of 3, passing the 3 test. Does not end in 0 or 5, passing the 5 test. Option A: 15 ends in 5, failing the 5 test. Option B: 21 has digit sum 3, failing the 3 test. Option C: 25 ends in 5, failing the 5 test. Only 11 passes all three tests.

Test each by digit sum: 61 has digit sum 6+1=7, not a multiple of 3. 62 has digit sum 6+2=8, not a multiple of 3. 63 has digit sum 6+3=9, a multiple of 3, so 63 is divisible by 3. Verified: 63 divided by 3 = 21 exactly. Option D: 64 has digit sum 10, not a multiple of 3.

Digit sum = 8 + 8 = 16. Since 16 is not a multiple of 3, the number 88 is not divisible by 3. Option A gives 14, option B gives 15, option D gives 17, none of which match the actual digit sum of 88.

The answer is True. 2 has exactly two factors, 1 and 2, satisfying the definition of a prime number. Every other even number is divisible by 2 in addition to 1 and itself, giving it at least three factors and making it composite. No other even number can be prime because 2 divides all of them.

49 is odd, digit sum 13 not a multiple of 3, does not end in 0 or 5, confirming A. 77 is odd, digit sum 14 not a multiple of 3, does not end in 0 or 5, confirming B. 91 is odd, digit sum 10 not a multiple of 3, does not end in 0 or 5, confirming C. Option D: 65 ends in 5, failing the 5 test immediately. Note that passing these checks does not guarantee primality — 49 = 7 times 7 and 91 = 7 times 13 are both composite.

75 ends in 5, failing the 5 test and confirming divisibility by 5. Also digit sum 7+5=12, a multiple of 3, confirming divisibility by 3 as well. Option A: 71 passes all three tests. Option B: 73 passes all three tests. Option D: 79 passes all three tests. Only 75 is immediately identified as composite.

The last digit of 246 is 6, which is even. Any number whose last digit is 0, 2, 4, 6, or 8 is divisible by 2. Verified: 246 divided by 2 = 123 exactly. Option A gives 4, option B gives 5, option D gives 8, none of which are the actual last digit of 246.

The answer is True. A number ending in 0 is even, confirming divisibility by 2. It also ends in 0, which satisfies the divisibility rule for 5. For example 30, 50, 70, and 100 all end in 0 and are all divisible by both 2 and 5. Numbers ending in 0 are in fact divisible by 10 = 2 times 5.

39: digit sum 3+9=12, a multiple of 3, confirming A. 66: digit sum 6+6=12, a multiple of 3, confirming C. 81: digit sum 8+1=9, a multiple of 3, confirming D. Option B: 43 has digit sum 4+3=7, not a multiple of 3, so 43 is not divisible by 3.

95 ends in 5, and any number ending in 0 or 5 is divisible by 5. Verified: 95 divided by 5 = 19 exactly. Option A: 92 ends in 2, not divisible by 5. Option B: 94 ends in 4, not divisible by 5. Option D: 97 ends in 7, not divisible by 5.

Any even number greater than 2 is composite because it has 2 as a factor. 46 ends in 6, making it even and immediately eliminated. Option B: 49 is odd. Option C: 51 is odd. Option D: 55 is odd. Only 46 fails the divisibility-by-2 test.

The answer is False. Passing the 2, 3, and 5 divisibility tests only makes a number a candidate, not a confirmed prime. For example 49 is not divisible by 2, 3, or 5, but 49 = 7 times 7 so it is composite. Similarly 77 = 7 times 11 passes all three tests but is not prime.

62 is even so divisible by 2, confirming A. 65 ends in 5 so divisible by 5, confirming B. 84 is even so divisible by 2, confirming D. Option C: 71 is odd and does not end in 0 or 5, so it passes both the 2 and 5 tests and is not immediately ruled out.

51 has digit sum 5+1=6, which is a multiple of 3, so 51 is divisible by 3 and eliminated. Option A: 34 has digit sum 7, not a multiple of 3. Option C: 62 has digit sum 8, not a multiple of 3. Option D: 77 has digit sum 14, not a multiple of 3. Only 51 is eliminated by the divisibility-by-3 test.

85 ends in 5, and any number ending in 0 or 5 is divisible by 5. Verified: 85 divided by 5 = 17 exactly. Option A: 85 is odd so not divisible by 2. Option B: digit sum 8+5=13, not a multiple of 3. Option D: 85 divided by 7 = 12 remainder 1, not divisible by 7.

The answer is True. An even number greater than 2 has at least three distinct factors: 1, 2, and itself divided by 2. Having more than two factors means it cannot be prime and must be composite. The only even prime is 2 itself, since it has exactly two factors: 1 and 2.

27: digit sum 2+7=9, a multiple of 3, confirming A. 42: digit sum 4+2=6, a multiple of 3, confirming C. 75: digit sum 7+5=12, a multiple of 3, confirming D. Option B: 34 has digit sum 3+4=7, not a multiple of 3, so 34 is not divisible by 3.

49 is odd so it passes the 2 test. Digit sum 4+9=13 is not a multiple of 3, passing the 3 test. It does not end in 0 or 5, passing the 5 test. Option B: 44 is even, failing the 2 test. Option C: 45 ends in 5 and has digit sum 9, failing both the 5 and 3 tests. Option D: 50 ends in 0, failing the 5 test.

Digit sum = 5 + 7 = 12. Since 12 is a multiple of 3, the number 57 is divisible by 3. Option A gives 10, option B gives 11, option D gives 13, none of which match the actual digit sum of 57.

The answer is True. The divisibility rule for 5 states that any number ending in 0 or 5 is a multiple of 5. For example 35, 50, 75, and 100 all end in 0 or 5 and are all divisible by 5. This rule works because our number system is base 10 and 10 = 2 times 5.