Number System Questions and Answers

Q: If a set of toys is arranged in rows of 20 each, 15 toys are left over; in rows of 25 each, 20 are left over; in rows of 38 each, 33 are left over; and in rows of 40 each, 35 are left over. What is the minimum number of toys?

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.

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121.

It is being given that \((2^{32} + 1)\) is completely divisible by a whole number. Which of the following numbers is completely divisible by this number?

A.\(2^{16} + 1\)
B.\(2^{16} - 1\)
C.\(7 \times 2^{23}\)
D.\(2^{96} + 1\)

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.

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122.

\(14927 \times 567 - 14927\times 467 = y\) . What is \(y\) ?

A.2985400
B.1492700
C.7463500
D.7483500

Answer: B

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

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123.

If a set of toys is arranged in rows of 20 each, 15 toys are left over; in rows of 25 each, 20 are left over; in rows of 38 each, 33 are left over; and in rows of 40 each, 35 are left over. What is the minimum number of toys?

A.1,255
B.3,805
C.7,595
D.3,795

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.

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124.

How many keystrokes are needed to type numbers from 1 to 1000 on a standard keyboard?

A.3001
B.2893
C.2704
D.2890

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.

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125.

What is the unit digit of \(13^{35}\) ?

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 \(7\)

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126.

When a number is divided by 17, the remainder is 9. When the same number is divided by 23, the remainder is 4. Find the number?

A.20
B.349
C.329
D.60

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 \(p = 20\)
\(x = 17p + 9\)
\(= 17 \times 20 + 9\)
\(= 349\)

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127.

The sum of the two digits of a number is 10. If the number is subtracted from the number obtained by reversing its digits, the result is 54. Find the number?

A.34
B.28
C.12
D.17

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 \(= 28 \)

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128.

How many ordered pairs of integer (x,y)(x,y) are there such that their product is a positive integer less than 100.

A.546
B.636
C.1090
D.946

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.

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129.

When 75% of a two-digit number is added to it, the digits of the number are reversed. Find the ratio of the unit's digit to the ten's digit in the original number.

A.1:2
B.1:4
C.2:1
D.3:2

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.

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130.

The natural numbers below 660 which are divisible by 5 and 11 but not by 3 are?

A.8
B.9
C.10
D.11

Answer: A

If the number is divisible by 5 and 11 it must be divisible by 55.

The numbers are less than 660.

Hence, dividing 659 by 55 gives the number of multiples of 55 = 11 (ignoring fraction part).

The 11 multiples of 55 which are less than 560, but of these 11 multiples some can be multiples of 3.

The numbers of such, multiples is the quotient of 11 by 3.
Quotient of 11/3=3.
Out of 11 multiples of 55, 3 are multiples of 3.

Hence, numbers less than 660 and divisible by 5 and 11 but not by 3=11−3=8

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