Number System Questions and Answers

Q: \((112 \times 5^{4}) = ?\)

\[ 5^{4} = 625 \]\[ 112 \times 625 = 112 \times \frac{10000}{16} = \frac{1{,}120{,}000}{16} = 70{,}000 \]Writing 625 as \( \tfrac{10^{4}}{16} \) turns the multiplication into a division by 16.

Q: \((935421 \times 625) = ?\)

Write 625 as \( \tfrac{10^{4}}{16} \).\[ 935421 \times 625 = \frac{9{,}354{,}210{,}000}{16} = 584{,}638{,}125 \]

Q: What least number must be added to 1056, so that the sum is completely divisible by 23 ?

Divide and look at the remainder.\[ 1056 = 23 \times 45 + 21 \]\[ 23 - 21 = 2 \]So adding 2 gives \( 1058 = 23 \times 46 \). Adding the remainder itself would make it divisible only by coincidence - it is the shortfall to the next multiple that is wanted.

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81.

If aa and bb are positive integers, and x=2×3×7×ax=2×3×7×a , and y=2×2×8×by=2×2×8×b , and the values of both xx and yy lie between 120 and 130 (not including the two), then a-b=a-b=

A.-2
B.-1
C.1
D.2

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

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82.

What is the remainder left after dividing 1!+2!+3!.....+100! by 7?

A.5
B.14
C.21

Answer: A

From \( 7! \) onward every factorial contains the factor 7, so all of them leave remainder 0.

Only the first six terms matter:

\[ 1 + 2 + 6 + 24 + 120 + 720 = 873 \]

\[ 873 = 7 \times 124 + 5 \]

So the remainder is 5.

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83.

x divides \(\left ( \left ( 49 \right )^{15} -1\right )\) completely. What is \(x\) ?

A.8
B.14
C.50
D.51

Answer: A

\(\left (x^{n}-1 \right )\) is divisible by \(x+1\) if \(x\) is an even number.
\(\left ( \left ( 49 \right )^{15} -1\right )\) =\(\left ( \left ( 7^{2} \right )^{15} -1\right )\) =\(\left ( 7 \right )^{30} -1\)
which is divisible by \(7+1\), \(8\)

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84.

The sum of three consecutive integers is 102. Find the lowest of the three?

A.40
B.53
C.29
D.33

Answer: D

Three consecutive numbers can be taken as \((P - 1), P, (P + 1)\)
So, \((P - 1) + P + (P + 1) = 102\)
\(3P = 102 => P = 34\)
The lowest of the three \(= (P - 1) = 34 - 1 = 33\)

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85.

\((112 \times 5^{4}) = ?\)

A.67000
B.70000
C.76500
D.77200

Answer: B

\[ 5^{4} = 625 \]

\[ 112 \times 625 = 112 \times \frac{10000}{16} = \frac{1{,}120{,}000}{16} = 70{,}000 \]

Writing 625 as \( \tfrac{10^{4}}{16} \) turns the multiplication into a division by 16.

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86.

\((935421 \times 625) = ?\)

A.575648125
B.584638125
C.584649125
D.585628125

Answer: B

Write 625 as \( \tfrac{10^{4}}{16} \).

\[ 935421 \times 625 = \frac{9{,}354{,}210{,}000}{16} = 584{,}638{,}125 \]

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87.

When writing numbers from 1 to 10,000, how many times is the digit 9 written ?

A.3200
B.3600
C.4000
D.4200

Answer: C

Treat every number from 0000 to 9999 as a four-digit string with leading zeros. That is 10,000 strings and 40,000 digit positions, each digit equally likely:

\[ \frac{40{,}000}{10} = 4000 \]

The padding adds only zeros, so the count of 9s is unaffected, and 10,000 itself contains no 9. So the digit 9 is written 4000 times.

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88.

The numerator of a certain fraction is 8 less than the denominator. If 3 is added to the numerator and 3 is subtracted from the denominator, the fraction becomes \(\dfrac{3}{4}\). Find the original fraction?

A.\(\dfrac{2}{5}\)
B.\(\dfrac{7}{9}\)
C.\(\dfrac{3}{11}\)
D.\(\dfrac{8}{5}\)

Answer: C

The denominator be \(P\), the numerator will be \((P - 8)\)
The fraction will be \(\dfrac{(P - 8)}{P}\)
Adding 3 to the numerator and subtracting 3 from the denominator,
\(\dfrac{(P - 8 + 3)}{P - 3}= \dfrac{3}{4}\)
\(\dfrac{(P - 5)}{P - 3}= \dfrac{3}{4}\)
\(P = 20 - 9 \Rightarrow P = 11\)
The fraction is: \(\dfrac{3}{11}\)

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89.

a,b,c,d and e are five consecutive numbers in increasing order of size. Deleting one number from the set decreased the sum of the remaining numbers in the set by 20%.

Which one of the following numbers was deleted?

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 \( c \).

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90.

What least number must be added to 1056, so that the sum is completely divisible by 23 ?

A.2
B.3
C.18
D.21
E.None of these

Answer: A

Divide and look at the remainder.

\[ 1056 = 23 \times 45 + 21 \]

\[ 23 - 21 = 2 \]

So adding 2 gives \( 1058 = 23 \times 46 \). Adding the remainder itself would make it divisible only by coincidence - it is the shortfall to the next multiple that is wanted.

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