GRE Quantitative Reasoning Practice: Number Properties

360 = 2³ × 3² × 5. Each divisor picks an exponent of 2 from 0 to 3, of 3 from 0 to 2 and of 5 from 0 to 1, so there are 4 × 3 × 2 = 24 divisors. Multiplying the exponents themselves (3 × 2 × 1) without adding 1 to each gives far too few.

84 = 2² × 3 × 7 and 120 = 2³ × 3 × 5. The LCM takes the highest power of each prime: 2³ × 3 × 5 × 7 = 840. The product 10,080 is a common multiple, but not the least one, and 12 is the greatest common divisor.

Write n = 6q + 5. Then 4n = 24q + 20, and 24q is a multiple of 6, so the remainder comes from 20: 20 = 18 + 2. Checking with n = 5 gives 20, remainder 2.

The primes between 20 and 40 are 23, 29, 31 and 37, and their sum is 120. Including 19, which is below 20, would give 139.

2n is always even, so 2n + 3 is always odd. n² + n = n(n + 1) is always even, and the others depend on n: 3n + 1 is even when n = 1.

The units digits of powers of 7 repeat in the cycle 7, 9, 3, 1. Since 82 = 4 × 20 + 2, 7⁸² ends like 7², which ends in 9.

There are 25 multiples of 4 and 16 multiples of 6. Numbers counted twice are the multiples of 12, of which there are 8, so the total is 25 + 16 − 8 = 33. Forgetting to subtract the overlap gives 41.

288 = 32 × 9 = 2⁵ × 3², so a = 5 and b = 2, and a + b = 7.

Among three consecutive integers at least one is even and exactly one is a multiple of 3, so the product is always divisible by 6. The example 1 × 2 × 3 = 6 shows that 8, 9, 12 and 24 do not always divide the product.

Each trailing zero needs a factor 10 = 2 × 5, and factors of 2 are plentiful, so count factors of 5. The multiples of 5 up to 30 give 6, and 25 contributes one extra, for 7. Stopping at 6 misses the second 5 inside 25.