Numbers Practice Test Questions - Quantitative Aptitude

2n is always even, so 2n + 1 is always odd. The other options can be even or odd depending on n.

A number divisible by both 3 and 11 is divisible by 33.
111,111: the digit sum = 6, so it is divisible by 3, and the alternate digit sums are 3 and 3 (difference 0), so it is divisible by 11. In fact 111,111 = 33 × 3367.
The other options are not divisible by both (111,111,111 has the alternate sums 5 and 4, so it is not divisible by 11; 1,111,111 and 11,111 have digit sums 7 and 5).

A number divisible by both 3 and 88 must be divisible by 3 and by 88.
888,888: the sum of digits = 48, so it is divisible by 3, and 888,888 ÷ 88 = 10,101, so it is divisible by 88.
The other options are not divisible by 3 or 88 (88,888,888 has the digit sum 64, 8,888,888 has 56, and 8,888 has 32).

In 12,345,678 the digits from the right are 8 (units), 7 (tens), 6 (hundreds), 5 (thousands) and 4 (ten-thousands).
Increasing the number by exactly 10,000 means increasing the ten-thousands digit, which is 4, by 1.

Three consecutive natural numbers are n, n + 1 and n + 2. Their sum = 3n + 3 = 3(n + 1), which is always divisible by 3.

20! contains the factors 2 and 3, so it is divisible by 6.
The remainder is 0.

The first six odd composite numbers are 9, 15, 21, 25, 27 and 33.
Sum = 130. Average = 1306 = 653.

For 8376$91# to be divisible by 11, the difference between the sum of the digits at odd places and at even places must be 0 or a multiple of 11.
Odd places: 8 + 7 + $ + 1 = 16 + $. Even places: 3 + 6 + 9 + # = 18 + #. Difference = $ - # - 2.
So $ - # = 2 or $ - # = -9. $ - # = 2 gives 8 pairs ($ = 2 to 9), and $ - # = -9 gives 1 pair ($ = 0, # = 9).
Total = 8 + 1 = 9.

999,999 = 7 × 142,857, so it is divisible by 7. Also 343343 = 343 × 1001 is divisible by 7, but it is smaller. 99 and 999 are not divisible by 7 (99 = 7 × 14 + 1 and 999 = 7 × 142 + 5).
The largest number divisible by 7 is 999,999.

4199 = 13 × 17 × 19 and 1547 = 7 × 13 × 17.
Let the integers in increasing order be a, b, c, d. Then bcd = 4199 and abc = 1547, so b and c are the common factors 13 and 17, d = 19 and a = 7.
Sum = 7 + 13 + 17 + 19 = 56.

Among the first 200 natural numbers there are 100 multiples of 2, 66 multiples of 3 and 33 multiples of 6 (multiples of both).
Multiples of exactly one of 2 or 3 = (100 - 33) + (66 - 33) = 67 + 33 = 100.
S withdraws these 100 slips, so 200 - 100 = 100 slips remain, which is not among the options.

330 = 2 × 3 × 5 × 11. For p! to be divisible by 330, the prime factor 11 must be present, so p ≥ 11.
11! contains 2, 3, 5 and 11, so the smallest p is 11.

The last digit of a square depends only on the last digit of the number. 3 × 3 = 9, so the square of a number ending in 3 ends in 9.

The even prime number is 2, so the sum of the 5 consecutive multiples of 8 is 2 - 2 = 0.
The middle multiple is the average, 05 = 0, so the multiples are -16, -8, 0, 8, 16.
The greatest multiple is 16.

178 = 13 × 13 + 9, so 178 leaves the remainder 9. 180 = 13 × 13 + 11, so 180 leaves the remainder 11.
9² = 81, and 81 = 6 × 13 + 3, so the remainder is 3.
11² = 121 = 9 × 13 + 4, so 11⁴ leaves the same remainder as 4² = 16, that is 3.
The remainder of the product = 3 × 3 = 9.

The product of four primes whose sum is 43: 4862 = 2 × 11 × 13 × 17 and 2 + 11 + 13 + 17 = 43.
The other options do not give a prime sum of 43: 1155 = 3 × 5 × 7 × 11 has the sum 26, and 1495 = 5 × 13 × 23 has only three factors.
So the product is 4862.

The divisor leaves a remainder 9, so it divides 2176 - 9 = 2167 and must be greater than 9.
2167 = 11 × 197, so among the options the prime number that divides 2167 is 197.

The first four prime numbers are 2, 3, 5 and 7.
Their sum = 2 + 3 + 5 + 7 = 17.

As decimals: 99101~0.9802, 9799~0.9798, 9597~0.9794, confirming 99101>9799>95/97.

As x grows large, x/(2x+1) approaches 1/2.

2n is always even regardless of n, so 2n+1 is always odd.

After removing 25, remaining = 38 × 35 = 22.8 L. After removing 35 of that, remaining = 22.8 × 25 = 9.12 L.

34=0.75 and 56~0.833. Among the options, only 45=0.8 lies strictly between them.

Total fruit needed = 60 × 12 = 30 cups. Cans needed = 303.5 = 8.57, rounded up to 9 whole cans.

All factors are odd, so the product has no factor of 2. Since a trailing zero requires a factor of 10 (2 × 5), there are no trailing zeroes.