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: B

There are 25 prime numbers between 1 and 100.

They are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89 and 97.

A quick check by decades: 4 primes below 10, then 4, 2, 2, 3, 2, 2, 3, 2 and 1 in the following tens, totalling 25. 26 usually comes from counting 1 as a prime, which it is not; 24 from missing 2, the only even prime.

Answer: A

There are 17 such numbers.

Multiples of 3 from 1 to 100: \(\lfloor 100/3 \rfloor = 33\). Those that are also multiples of 2 are the multiples of 6: \(\lfloor 100/6 \rfloor = 16\).

\[ 33 - 16 = 17 \]

These are the odd multiples of 3: 3, 9, 15, ..., 99. 34 is roughly double, and 21 and 22 fit no step.

Answer: B

There are 128 three-digit numbers divisible by 7.

The smallest three-digit multiple of 7 is \(105 = 7 \times 15\) and the largest is \(994 = 7 \times 142\). Count the multipliers from 15 to 142:

\[ 142 - 15 + 1 = 128 \]

Another route: there are \(\lfloor 999/7 \rfloor = 142\) multiples of 7 up to 999 and \(\lfloor 99/7 \rfloor = 14\) up to 99, and \(142 - 14 = 128\). 129 is the off-by-one error of counting 14 to 142, and 126 or 131 come from misjudging the first or last multiple.

Answer: C

The numerator is 3.

To change the denominator from 4 to 12, multiply top and bottom by 3:

\[ \begin{aligned} \frac{1}{4} &= \frac{1 \times 3}{4 \times 3} \\ &= \frac{3}{12} \end{aligned} \]

8 is what you get by subtracting 4 from 12 instead of dividing, and the negative options would change the sign of a positive fraction, which multiplying top and bottom by the same number never does.

Answer: B

The other number is \(-\frac{7}{15}\).

Subtract the known number from the sum, writing both over 15:

\[ \begin{aligned} -\frac{3}{5} - \left(-\frac{2}{15}\right) &= -\frac{9}{15} + \frac{2}{15} \\ &= -\frac{7}{15} \end{aligned} \]

Check: \(-\frac{7}{15} + \left(-\frac{2}{15}\right) = -\frac{9}{15} = -\frac{3}{5}\). \(\frac{7}{15}\) has the wrong sign, and \(-\frac{8}{15}\) or \(\frac{8}{5}\) come from mishandling the minus signs or the common denominator.

Answer: C

The greatest three-digit number divisible by 7 is 994.

Divide the greatest three-digit number, 999, by 7:

\[ \begin{aligned} 999 &= 7 \times 142 + 5 \\ 999 - 5 &= 994 = 7 \times 142 \end{aligned} \]

998 and 976 leave remainders when divided by 7. 987 is divisible by 7 (\(7 \times 141\)), but it is smaller than 994.

The same method works for any question of this type: divide the largest candidate by the divisor and subtract the remainder.

Answer: B

The product is 25000.

In 65231 the digit 5 is in the thousands place, so its place value is 5000. Its face value is the digit itself, 5.

\[ 5000 \times 5 = 25000 \]

The other options come from taking the wrong place value or adding instead of multiplying; for example, 5000 + 5 is 5005, which is not offered.

The face value of a digit never depends on its position, while its place value is the digit multiplied by the value of its place (here 5 × 1000).

Answer: C

\[ x = 9548 + 7314 - 8362 \]

\[ = 16862 - 8362 = 8500 \]