Number System Questions and Answers

Answer: B

We are given that x=2×3×7×a=42ax=2×3×7×a=42a and y=2×2×8×b=32by=2×2×8×b=32b

We are given that the values of both xx and yy lie between 120 and 130 (not including the two).

The only multiple of 42 in this range is 42×3=12642×3=126.

Hence, x=126x=126 and a=3.a=3.
The only multiple of 32 in this range is 32×4=12832×4=128.

Hence, y=128y=128 and b=4b=4.

Hence, a−b=3−4=a−b=3−4= -1

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: A

The last possible card number is 996, and the useful observation is that this is just short of the modulus:

\[ 996 \equiv -1 \pmod{997} \]

The rule Pinaki wrote down says the product leaves remainder 123, so with \( P \) for the PIN

\[ 996 \times P \equiv 123 \pmod{997} \quad\Longrightarrow\quad -P \equiv 123 \pmod{997} \]

\[ P \equiv -123 \equiv 997 - 123 = 874 \pmod{997} \]

Check: \( 996 \times 874 = 870{,}504 \) and \( 870{,}504 = 997 \times 873 + 123 \). Replacing 996 by \( -1 \) turns a six-figure multiplication into a subtraction.

Answer: C

Let the divisor be dd. 

When 242 is divided by the divisor, let the quotient be 'xx' and we know that the remainder is 8.
Therefore, 242=xd+8242=xd+8

Similarly, let yy be the quotient when 698 is divided by dd.
Then, 698=yd+9698=yd+9.

242+698=940=xd+yd+8+9242+698=940=xd+yd+8+9
940=xd+yd+17940=xd+yd+17

As xd and ydxd and yd are divisible by dd, the remainder when 940 is divided by dd should have been 17.

However, as the question states that the remainder is 4, it would be possible only when 17/d leaves a remainder of 4.

If the remainder obtained is 4 when 17 is divided by dd, then dd has to be 13

Answer: A

We have to omit four digits in such a manner that we get the largest possible number as the result.

We will omit 2,1,32,1,3 and 55 and the result will be 98769876.

The largest omitted digit will be 5.

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: C

Five consecutive numbers are symmetric about the middle one, so their total is five times that middle value. Writing the set as \( c-2,\, c-1,\, c,\, c+1,\, c+2 \):

\[ S = 5c \]

Removing one member cuts the total by 20 per cent, so the number removed is

\[ 0.20 \times 5c = c \]

which is the middle number, c.

The result holds whatever the five numbers are, since the arithmetic never uses their actual values - only that they are consecutive and so centred on \( c \).

Answer: D

Ten cigarettes give 10 stub.

From 10 stubs 3 more cigarettes can be made (2 stubs would be obtained from 2 cigarettes formed by joining 8 stubs.)

So he could smoke 13 cigarettes in all.

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: C

We are given that the numbers mm and nn, when divided by 6, leave remainders of 2 and 3, respectively.

Hence, we can represent the numbers mm and nn as 6p+26p+2 and 6q+36q+3, respectively, where pp and qq are suitable integers.

Now,
m−n=(6p+2)−(6q+3)

=6p−6q−1

=6(p−q)−1

A remainder must be positive, so let’s add 6 to this expression and compensate by subtracting 6:
6(p−q)−1=6(p−q)−6+6−1

=6(p−q)−6+5

=6(p−q−1)+5

Thus, the remainder is 5