Number System Questions and Answers

Answer: C

From 100 to 200 there are 101 numbers.
There are 100, 1's in the hundred place.
10, 1's in tens place and10, 1's in unit place.

Thus, the answer is 100+10+10=100+10+10= 120.

Answer: B

Since each one of the three numbers aa, bb, and cc is divisible by 3, the numbers can be represented as 3p,3qand 3r respectively, where p,q and r are integers.

The product of the three numbers is 3p×3q×3r=27(pqr).
Since p,qp,q and rr are integers, pqrpqr is an integer and therefore abcabc is divisible by 27.

Answer: D

Count the 8s by the place they occupy.

Units digit 8 - the numbers 268, 278, …, 488:

\[ \frac{488 - 268}{10} + 1 = 23 \]

Tens digit 8 - the blocks 280–289, 380–389 and 480–489:

\[ 3 \times 10 = 30 \]

No number in the range has 8 in the hundreds place, since the hundreds digits are only 2, 3 and 4.

\[ 23 + 30 = 53 \]

288 and 388 and 488 are each counted twice - once for each 8 - which is correct, since the question asks how many 8s are written.

Answer: C

A number is divisible by 11, if the difference between the sum of digits at even places and odd places is either 0 or divisible by 11.

Numbers 5,3,6,4,7 is a multiple of 11

∵(5+6+7)−(3+4)=11, which is a multiple of 11

The number at odd places 5,6,7 can be arranged in 3! ways and 3,4 can be arranged in 2! ways.

Therefore, the total number of ways of such numbers =3!×2!=12

Answer: B

Test each candidate against both conditions, since the second is far more restrictive than it looks:

The pattern behind it: dividing by 14 leaves either \( x \) or \( x+7 \), and the second happens exactly when the quotient by 7 is odd. Here \( 53 = 7 \times 7 + 4 \), and that quotient 7 is odd, so the remainder on 14 is \( 4 + 7 = 11 \).

Answer: B

Every one of the nine numbers is a multiple of 111:

\[ 111,\ 222 = 2 \times 111,\ \ldots,\ 999 = 9 \times 111 \]

So any common divisor of all nine must divide 111, and conversely every divisor of 111 divides them all.

\[ 111 = 3 \times 37 \]

The divisors greater than 1 are therefore 3, 37 and 111. 9 is excluded because \( 111 \div 9 \) is not whole.

Answer: A

Any prime number greater than 3 will be in the form of 6x+1 or 6x−1.

Thus, both prime number are twins:

Let first be 6x−1
and 2nd be 6x+1
Sum=12x

Thus it is always divisible by 12.

Answer: A

From \( 7! \) onward every factorial contains the factor 7, so all of them leave remainder 0.

Only the first six terms matter:

\[ 1 + 2 + 6 + 24 + 120 + 720 = 873 \]

\[ 873 = 7 \times 124 + 5 \]

So the remainder is 5.

Answer: D

Count the factors of 3 first, then pair them up. Legendre's rule gives the exponent of a prime in a factorial:

\[ \left\lfloor\frac{99}{3}\right\rfloor + \left\lfloor\frac{99}{9}\right\rfloor + \left\lfloor\frac{99}{27}\right\rfloor + \left\lfloor\frac{99}{81}\right\rfloor = 33 + 11 + 3 + 1 = 48 \]

So \( 3^{48} \) divides 99! and no higher power does. Because \( 9 = 3^{2} \), each factor of 9 consumes two of those threes:

\[ \text{highest power of } 9 = \left\lfloor\frac{48}{2}\right\rfloor = 24 \]

Answer 48 is the trap for stopping at the power of 3 without halving it.

Answer: D

Work outward from the second statement.

\[ b = 3 \times 2 + 1 = 7 \]

\[ a = 3b + 2 = 3 \times 7 + 2 = 23 \]

Checking: \( 23 \div 3 = 7 \) remainder 2, and \( 7 \div 3 = 2 \) remainder 1.