Number System Questions and Answers

Answer: B

On dividing the given number by 342, let kk be the quotient and 47 as remainder.

Then,

number−342k+47=(19×18k+19×2+9)number−342k+47=(19×18k+19×2+9)

=19(18k+2)+9=19(18k+2)+9

⇒ The given number when divided by 19, gives 18k+218k+2 as quotient and 9 as remainder.

Answer: D

Let \(2^{32} = x\).
Then, \((2^{32} + 1) = (x + 1)\).
Let \((x + 1)\) be completely divisible by the natural number N. Then,
\((2^{96} + 1) \\= [(2^{32})^{3} + 1]\\= (x^{3} + 1)\\= (x + 1)(x^{2} - x + 1)\)
which is completely divisible by N, since \((x + 1)\) is divisible by N.

Answer: B

\(y = 14927 \times 567 - 14927 \times 467\)
\(= 14927 \times (567 - 467)\)
\(= 14927 \times 100\)
\(= 1492700\)

Answer: D

In every case the shortfall is the same:

\[ 20 - 15 = 25 - 20 = 38 - 33 = 40 - 35 = 5 \]

So \( N + 5 \) is divisible by all four numbers.

\[ \text{LCM}(20, 25, 38, 40) = 2^{3} \times 5^{2} \times 19 = 3800 \]

\[ N = 3800 - 5 = 3795 \]

Spotting the common shortfall is what turns four separate congruences into a single LCM.

Answer: B

While typing numbers from 1 to 1000, you have 9 single digit numbers from 1 to 9. Each of them requires one keystroke. That is 9 key strokes.

There are 90 two-digit numbers, from 10 to 99. Each of these numbers requires 2 keystrokes. Therefore, one requires 180 keystrokes to type the 2 digit numbers.

There are 900 three-digit numbers, from 100 to 999. Each of these numbers requires 3 keystrokes. Therefore, one requires 2700 keystrokes to type these 3 digit numbers.

Then 1000 is a four-digit number which requires 4 keystrokes.

Totally, therefore, one requires 9+180+2700+4=9+180+2700+4= 2893 keystrokes.

Answer: D

\(13^{1}\) has unit digit \(3\)
\(13^{2}\) has unit digit \(9\)
\(13^{3}\) has unit digit \(7\)
\(13^{4}\) has unit digit \(1\)
This will form a pattern \(3-7-9-1\)
So, \(13^{32}\) has unit digit \(1\)
\(13^{35}\) has unit digit as \(7\)

Answer: B

\(x = 17p + 9\) and \(x = 23q + 4\)
i.e. \(17p + 9 = 23q + 4\)
Therefore, \(q = \frac{(17p + 5)}{23}\)
Least value of p for which q is a whole number is \(p = 20\)
\(x = 17p + 9\)
\(= 17 \times 20 + 9\)
\(= 349\)

Answer: B

Any two digit number can be written as \((10P + Q)\), where \(P\) is the digit in the tens place and \(Q\) is the digit in the units place.
\(P + Q = 10\) ----- (1)
\((10Q + P) - (10P + Q) = 54\)
\(9(Q - P) = 54\)
\((Q - P) = 6\) ----- (2)
Solve (1) and (2) \(P = 2\) and \(Q = 8\)
The required number is \(= 28 \)

Answer: D

Given 0 x and y are either both positive or both negative.

Also given (x,y) is not equal to (y,x).
If x=1, y can take values from 1 to 99
⇒ we have 99×2=198 pairs but (1,1) is repeated
Thus can take 198−1=197 pairs.

If x=2 , yy can take values from 2 to 49
[(2,1) and (1,2) are also covered in 197 pairs above].
⇒ 48×2−1=95 pairs [(2,2) is repeated]

Similarly,
If x=3 or y=3 we have 61 pairs
If x=4 or y=4 we have 41 pairs
If x=5 or y=5 we have 29 pairs
If x=6 or y=6 we have 21 pairs
If x=7 or y=7 we have 15 pairs
If x=8 or y=8 we have 9 pairs
If x=9 or y=9 we have 5 pairs

We have total 473 pairs when x and y are positive.
We will have 473 pairs when a and b are negative.
⇒ We have a total of 946 ordered pairs.

Answer: C

Let the number be \( 10t + u \), with \( t \) the tens digit and \( u \) the units digit. Adding 75 per cent of it multiplies it by 1.75, and the result is the reversed number:

\[ 1.75\left(10t + u\right) = 10u + t \]

\[ 17.5t + 1.75u = 10u + t \]

\[ 16.5t = 8.25u \quad\Longrightarrow\quad \frac{u}{t} = \frac{16.5}{8.25} = 2 \]

So the units digit is twice the tens digit, a ratio of 2 : 1. The smallest number fitting is 12, and indeed \( 12 + 0.75(12) = 21 \).

Read the order carefully: the question asks unit to ten, so 1 : 2 is the same relationship stated backwards.