Number System Questions and Answers

Q: If x is a whole number, then \(x^{2}(x^{2}-1)\) is always divisible by:

Factorise the expression.\[ x^{2}(x^{2}-1) = (x-1)\,x\,x\,(x+1) \]\( (x-1)x(x+1) \) is a product of three consecutive integers, so it is divisible by both 2 and 3, hence by 6; the extra factor of \( x \) supplies a second factor of 2 whenever \( x \) is even, and when \( x \) is odd both neighbours are even.Either way the product is divisible by 12. It is not always divisible by 24: \( x = 2 \) gives 12.

Q: The number 3 divides ‘a’ with a result of ‘b’ and a remainder of 2. The number 3 divides ‘b’ with a result of 2 and a remainder of 1. What is the value of a?

Work outward from the second statement.\[ b = 3 \times 2 + 1 = 7 \]\[ a = 3b + 2 = 3 \times 7 + 2 = 23 \]Checking: \( 23 \div 3 = 7 \) remainder 2, and \( 7 \div 3 = 2 \) remainder 1.

Q: A bank issued credit card numbers and the corresponding PIN (Personal Identification Number). Both are 3-digit numbers up to 996. Pinaki was the last to get the credit card and so he had the last possible credit card number. He was afraid of forgetting his PIN. He wrote down the number 123 in his diary to remember his PIN. He also wrote out the way to calculate 123 : "Multiply the card number by PIN. Divide the product by 997. The remainder is 123". Once, Prafull saw his diary in which Pinaki wrote this number 123. Prafull did a lot of purchasing, as he now knows Pinaki's PIN. What is Pinaki's PIN?

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.

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131.

If x is a whole number, then \(x^{2}(x^{2}-1)\) is always divisible by:

A.12
B.24
C.12-x
D.multiple of 12

Answer: A

Factorise the expression.

\[ x^{2}(x^{2}-1) = (x-1)\,x\,x\,(x+1) \]

\( (x-1)x(x+1) \) is a product of three consecutive integers, so it is divisible by both 2 and 3, hence by 6; the extra factor of \( x \) supplies a second factor of 2 whenever \( x \) is even, and when \( x \) is odd both neighbours are even.

Either way the product is divisible by 12. It is not always divisible by 24: \( x = 2 \) gives 12.

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132.

The number 3 divides ‘a’ with a result of ‘b’ and a remainder of 2.

The number 3 divides ‘b’ with a result of 2 and a remainder of 1.

What is the value of a?

A.13
B.17
C.21
D.23

Answer: D

Work outward from the second statement.

\[ b = 3 \times 2 + 1 = 7 \]

\[ a = 3b + 2 = 3 \times 7 + 2 = 23 \]

Checking: \( 23 \div 3 = 7 \) remainder 2, and \( 7 \div 3 = 2 \) remainder 1.

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133.

The sum of three prime numbers is 100. If one of them exceeds another by 36, then one of the numbers is:

A.7
B.29
C.41
D.67

Answer: D

\(x+(x+36)+y=100 \\ \Rightarrow 2x+y=64\)
Therefore \(y\) must be even prime, which is 2.
Therefore,
\(2x+2=64 \\ \Rightarrow x=31\)
Third prime number
\(= (x+36)\\= (31+36)\\= 67\)

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134.

A set of S consists of,

i). All odd numbers from 1 to 55.
ii). All even numbers from 56 to 150.

What is the index of the highest power of 3 in the product of all the elements of the set SS?

A.35
B.48
C.6
D.36

Answer: A

Only the multiples of 3 in the set contribute, and each contributes as many threes as it contains.

Odd numbers from 1 to 55. The odd multiples of 3 are 3, 9, 15, 21, 27, 33, 39, 45, 51, carrying exponents 1, 2, 1, 1, 3, 1, 1, 2, 1:

\[ 1+2+1+1+3+1+1+2+1 = 13 \]

Even numbers from 56 to 150. These are the multiples of 6 in the range, 60 through 150, sixteen of them, with exponents 1, 1, 2, 1, 1, 2, 1, 1, 3, 1, 1, 2, 1, 1, 2, 1:

\[ 1+1+2+1+1+2+1+1+3+1+1+2+1+1+2+1 = 22 \]

\[ \therefore\; \text{index} = 13 + 22 = 35 \]

Note that 27 and 108 each supply three, and 9, 45, 72, 90, 126 and 144 each supply two; counting one per multiple would give 25 and miss the higher powers.

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135.

A bank issued credit card numbers and the corresponding PIN (Personal Identification Number). Both are 3-digit numbers up to 996. Pinaki was the last to get the credit card and so he had the last possible credit card number.

He was afraid of forgetting his PIN. He wrote down the number 123 in his diary to remember his PIN. He also wrote out the way to calculate 123 : "Multiply the card number by PIN. Divide the product by 997. The remainder is 123".

Once, Prafull saw his diary in which Pinaki wrote this number 123. Prafull did a lot of purchasing, as he now knows Pinaki's PIN. What is Pinaki's PIN?

A.874
B.875
C.876
D.877

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.

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136.

A chain smoker had spent all the money he had. He had no money to buy his cigarettes. Hence, he resorted to join the stubs and to smoke them.

He needed 4 stubs to make a single cigarette. If he got a pack of 10 cigarettes as a gift, then how many cigarettes could he smoke in all?

A.10
B.11
C.12
D.13

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.

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137.

In a meet, people from five different places have assembled in Bangalore High School. From the five places the persons come to represent are 42,60,210,90 and 84.

What is the minimum number of rooms that would be required to accommodate so that each room has the same number of occupants and occupants are all from the same places?

A.44
B.62
C.81
D.96

Answer: C

All the students from each have to be accommodated in a certain number of rooms.

There should be no person left over (remainder) from any places who can be clubbed together with the persons left over from other places.

To have the minimum number of rooms, the capacity of each room is HCF of all the numbers.
HCF(42,60,210,90,84)=6HCF(42,60,210,90,84)=6

Thus Min number of rooms =(42+60+210+90+84) / 6=81

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