Number System Questions and Answers

Answer: A

There are 7 such numbers.

List the multiples of 4 between 1 and 100 and keep those with a digit 4:

4, 24, 40, 44, 48, 64, 84

Numbers with a digit 4 such as 14, 34, 54 or 94, and 41, 42, 43, 45, 46, 47 and 49 in the forties, are not multiples of 4, so they do not count. The count is 7. Answers of 20 or 21 come from counting every number with a digit 4 (there are 19 between 1 and 100) or every multiple of 4 (25), without requiring both conditions.

Answer: A

The digit 7 is written 20 times.

Count the units place and the tens place separately from 1 to 100:

Units digit 7: 7, 17, 27, ..., 97, which is 10 times.
Tens digit 7: 70, 71, ..., 79, which is 10 times.

That makes 20 in all. The number 77 contributes one 7 to each count, which is correct, since 7 is written twice in it. 19 forgets that, and 21 or 22 count 77 or 70 to 79 one time too many.

Answer: C

\[ x = 9548 + 7314 - 8362 \]

\[ = 16862 - 8362 = 8500 \]

Answer: D

The decimal expansion sorts the real numbers into two kinds:

• Rational - the expansion either stops or falls into a repeating block, as in \( \tfrac{1}{4} = 0.25 \) or \( \tfrac{1}{3} = 0.333\ldots \)
• Irrational - it runs on for ever without ever repeating, as in \( \sqrt{2} = 1.41421356\ldots \) or \( \pi \)

A non-terminating, non-repeating expansion is therefore irrational. Natural and whole numbers are particular rationals with expansions that stop at once, so none of the other options can apply.

Answer: A

Divide the largest four-digit number by 88 and drop the remainder.

\[ 9999 = 88 \times 113 + 55 \]

\[ 9999 - 55 = 9944 = 88 \times 113 \]

9988 on the list is even and looks plausible, but \( 9988 \div 88 = 113.5 \).

Answer: C

Rather than computing the power in full, reduce the base first. Since \( 3^{2} = 9 \equiv 1 \pmod 8 \), every even power of 3 is 1 modulo 8:

\[ 3^{7} = \left(3^{2}\right)^{3} \cdot 3 \equiv 1^{3} \cdot 3 \equiv 3 \pmod 8 \]

The direct route agrees: \( 3^{7} = 2187 \) and \( 2187 = 8 \times 273 + 3 \).

Spotting that the base squares to 1 is what makes this quick, and the same trick handles any power of 3 modulo 8: odd powers leave 3, even powers leave 1.

Answer: B

Ten's digit = 7 ⇒ units digit =8 ⇒ Hundred's digit =1,2,3,4,5,6.
⇒ Number of ways =1×6

Ten's digit = 6 ⇒ units digit =7,8 ⇒ Hundred's digit =1,2,3,4,5.
⇒ Number of ways =2×5

Ten's digit = 5 ⇒ units digit =6,7,8 ⇒ Hundred's digit =1,2,3,4
⇒ Number of ways =3×4

Ten's digit = 4 ⇒ units digit =5,6,7,8 ⇒ Hundred's digit =1,2,3.
⇒ Number of ways =4×3

Ten's digit = 3 ⇒ units digit =4,5,6,78 ⇒ Hundred's digit =1,2
⇒ Number of ways =5×2

Ten's digit = 2 ⇒ units digit =3,4,5,6,7,8 ⇒ Hundred's digit =1
⇒ Number of ways =6×1

Total number of ways =6+10+12+12+10+6= 56.

Answer: C

We have given that the number 2ab52ab5 is divisible by 25.

Any number divisible by 25 ends with the last two digits 00, 25, 50, or 75.
So, b5b5 should equal 25 or 75.

Hence, b=2b=2 or 77.
Since aa is now free to take any digit from 0 through 9, abab can have multiple values.

We also have that abab is divisible by 13.
The multiples of 13 are 13,26,39,52,65,7813,26,39,52,65,78 and 491491.

Among these, the only number ending with 2 or 7 is 52.

Hence, ab=52

Answer: A

Every piece must weigh the same and be as heavy as possible, which is exactly the highest common factor of the three coins:

\[ 780 = 2^{2}\cdot3\cdot5\cdot13, \quad 840 = 2^{3}\cdot3\cdot5\cdot7, \quad 960 = 2^{6}\cdot3\cdot5 \]

\[ \operatorname{HCF} = 2^{2}\cdot3\cdot5 = 60 \text{ g} \]

Counting the pieces each coin yields:

\[ \frac{780}{60} + \frac{840}{60} + \frac{960}{60} = 13 + 14 + 16 = 43 \]

Each piece is shared by two people, so

\[ 43 \times 2 = 86 \text{ persons} \]

43 is the trap for stopping at the piece count and forgetting the sharing.

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.