Number System Questions and Answers

Q: The difference between the local value and face value of 7 in the numeral 657903 is:

The local (place) value of a digit is the digit multiplied by its place; the face value is the digit itself.In 657903 the 7 sits in the thousands place:\[ 7000 - 7 = 6993 \]

Q: A and B are playing mathematical puzzles. A asks B "the whole numbers, greater than one, which can divide all the nine three digit numbers 111,222,333,444,555,666, 777,888 and 999?" B immediately gave the desired answer. It was:

Every one of the nine numbers is a multiple of 111:\[ 111,\ 222 = 2 \times 111,\ \ldots,\ 999 = 9 \times 111 \]So any common divisor of all nine must divide 111, and conversely every divisor of 111 divides them all.\[ 111 = 3 \times 37 \]The divisors greater than 1 are therefore 3, 37 and 111. 9 is excluded because \( 111 \div 9 \) is not whole.

Q: Which one of the following is not a prime number?

Test each for divisibility.\[ 91 = 7 \times 13 \]31, 61 and 71 have no divisor other than 1 and themselves, so 91 is the composite one. It is a classic trap: it is odd, does not end in 5, and its digits sum to 10, so the easy tests all pass - only 7 catches it.

Q: A number when divided by the sum of 555 and 445 gives two times their difference as quotient and 30 as the remainder. The number is:

\[ \text{divisor} = 555 + 445 = 1000 \]\[ \text{quotient} = 2(555 - 445) = 220 \]\[ \text{number} = 1000 \times 220 + 30 = 220{,}030 \]

Important Instructions

91.

The remainder when the positive integer m is divided by 7 is x. The remainder when m is divided by 14 is x+7.

Which one of the following could m equal?

A.45
B.53
C.72
D.85

Answer: B

Test each candidate against both conditions, since the second is far more restrictive than it looks:

m	m mod 7 = x	m mod 14	x + 7	fits?
45	3	3	10	no
53	4	11	11	yes
72	2	2	9	no
85	1	1	8	no

The pattern behind it: dividing by 14 leaves either \( x \) or \( x+7 \), and the second happens exactly when the quotient by 7 is odd. Here \( 53 = 7 \times 7 + 4 \), and that quotient 7 is odd, so the remainder on 14 is \( 4 + 7 = 11 \).

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92.

The difference between the local value and face value of 7 in the numeral 657903 is:

A.7896
B.6993
C.903
D.1662

Answer: B

The local (place) value of a digit is the digit multiplied by its place; the face value is the digit itself.

In 657903 the 7 sits in the thousands place:

\[ 7000 - 7 = 6993 \]

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93.

A and B are playing mathematical puzzles. A asks B "the whole numbers, greater than one, which can divide all the nine three digit numbers 111,222,333,444,555,666, 777,888 and 999?"

B immediately gave the desired answer. It was:

A.3, 37 and 119
B.3, 37 and 111
C.9, 37 and 111
D.3, 9 and 37

Answer: B

Every one of the nine numbers is a multiple of 111:

\[ 111,\ 222 = 2 \times 111,\ \ldots,\ 999 = 9 \times 111 \]

So any common divisor of all nine must divide 111, and conversely every divisor of 111 divides them all.

\[ 111 = 3 \times 37 \]

The divisors greater than 1 are therefore 3, 37 and 111. 9 is excluded because \( 111 \div 9 \) is not whole.

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94.

Which one of the following is not a prime number?

A.31
B.61
C.71
D.91

Answer: D

Test each for divisibility.

\[ 91 = 7 \times 13 \]

31, 61 and 71 have no divisor other than 1 and themselves, so 91 is the composite one. It is a classic trap: it is odd, does not end in 5, and its digits sum to 10, so the easy tests all pass - only 7 catches it.

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95.

The positive integers mm and nn leave remainders of 2 and 3, respectively, when divided by 6. m>nm>n.

What is the remainder when m-nm-n is divided by 6?

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

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96.

1397 x 1397 = ?

A.1951609
B.1981709
C.18362619
D.2031719
E.None of these

Answer: A

\(1397\times 1397 = (1397)^{2}\)

\(= (1400-3)^{2}\)

\(= (1400)^{2}+(3)^{2}-(2\times 1400\times 3)\)

\(= 1960009 - 8400\)

\(= 1951609\)

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97.

\(\left \{(476 + 424)^{2} - 4 \times 476 \times 424 \right \} = ?\)

A.2906
B.3116
C.2704
D.2904
E.None of these

Answer: C

Given Expression = \([(a + b)^{2} - 4ab]\), where \(a = 476\) and \(b = 424\)
\(=[(476 + 424)^{2} - 4 \times 476 \times 424]\)
\(= [(900)^{2} - 807296]\)
\(= 810000 - 807296\)
\(= 2704\)

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98.

A number when divided by the sum of 555 and 445 gives two times their difference as quotient and 30 as the remainder. The number is:

A.1220
B.1250
C.22030
D.220030

Answer: D

\[ \text{divisor} = 555 + 445 = 1000 \]

\[ \text{quotient} = 2(555 - 445) = 220 \]

\[ \text{number} = 1000 \times 220 + 30 = 220{,}030 \]

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99.

Four digits of the number 29138576 are omitted so that the largest result is possible. The largest omitted digit is:

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.

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100.

The highest power of 9 dividing 99! completely is

A.12
B.14
C.22
D.24

Answer: D

Count the factors of 3 first, then pair them up. Legendre's rule gives the exponent of a prime in a factorial:

\[ \left\lfloor\frac{99}{3}\right\rfloor + \left\lfloor\frac{99}{9}\right\rfloor + \left\lfloor\frac{99}{27}\right\rfloor + \left\lfloor\frac{99}{81}\right\rfloor = 33 + 11 + 3 + 1 = 48 \]

So \( 3^{48} \) divides 99! and no higher power does. Because \( 9 = 3^{2} \), each factor of 9 consumes two of those threes:

\[ \text{highest power of } 9 = \left\lfloor\frac{48}{2}\right\rfloor = 24 \]

Answer 48 is the trap for stopping at the power of 3 without halving it.

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