Quant - Number System Ex. 3

Exercise3 + Answers + Explanations

Question:
When 26854 and 27584 are divided by a certain two digit prime number, the remainder obtained is 47. Which of the following choices is a possible value of the divisor?

(1) 61
(2) 71
(3) 73
(4) 89

Correct Answer - (3)

Explanation:
When both 26854 and 27584 are divided by a divisor, the remainder obtained happens to be the same value 47.

Which means that the difference between 26854 and 27584 is a multiple of the divisor. Else the remainders would have been different.

=> 27584 - 26854 = 730

Now we need to find out a two digit prime number, whose multiple will be 730. 730 can be written as 73 * 10, where 73 is a prime number and greater than 47.
Hence, the two digit prime divisor is 73.

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Question:
How many times will the digit ‘0′ appear between 1 and 10,000?

(1) 4000
(2) 4003
(3) 2893
(4) 3892
Correct Answer - (3)

Explanation:

In two digit numbers, ‘0′ appears 9 times.
In 3-digit numbers, either 1 ‘0′ or 2 ‘0s’ can appear.
When one ‘0′ appears: The first digit of a 3-digit number can be any of the 9 digits other than ‘0′. The second digit can be any of the ‘9′ digits and the third digit be ‘0′.Hence there will 9*9 = 81 such combinations. The ‘0′ can either be the 2nd digit or the third digit, hence there are 2 arrangements possible.Therefore, there are a total of 81 * 2 = 162 three digit numbers with ‘0′ as one of the digits.

When two ‘0s’ appear: The first digit of the number can be any of the 9 digits. The second and third digit are 0s.
Hence there are 9 such numbers and 0 appears 9*2 = 18 times. Hence, ‘0′ appears 18 + 162 = 180 times in three digit numbers. In 4-digit numbers, 0 might be either 1 or 2 or 3 of the digits.

When one ‘0′ appears: The remaining three digits can be any of the 9 digits.i.e. 93 combinations = 729. The ‘0′ can either be the second or third or fourth digit. Hence 3*729 numbers = 2187.

When two 0s appear: The remaining two digits can be any of the 9 digits.Hence 92 = 81 possible combinations. The two 0s can be in two of the three positions = 3*81 = 243 such numbers and 2*243 = 486 number of times 0 will be appear.

When three 0s appear: There are 9 such numbers and 0 appears thrice in each. Hence 9*3 = 27 times.
Hence the number of 0s in 4-digit numbers = 2187 + 486 + 27 = 2700 times.
There is one 5-digit number - 10,000 and 0 appears four times in this number.
Hence the total number of times 0 appears between 1 and 10,000 = 9 + 180 + 2700 + 4 = 2893.

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Question:
What number should be subtracted from x^3 + 4x^2 – 7x + 12 if it is to be perfectly divisible by x + 3?

(1) 42
(2) 39
(3) 13
(4) None of these
Correct Answer - (1)

Explanation:
According to remainder theorem when, then the remainder is f(-a). In this case, as x + 3 divides x3 + 4×2 – 7x + 12 – k perfectly (k being the number to be subtracted), the remainder is 0 when the value of x is substituted by –3.

=> (-3)3 + 4(-3)2 – 7(-3) + 12 – k = 0
=> -27 + 36 + 21 + 12 = k
=> k = 42

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Question:
What is the value of M and N respectively? If M39048458N is divisible by 8 and 11; Where M and N are single digit integers?

(1) 7, 8
(2) 8, 6
(3) 6, 4
(4) 5, 4

Correct Answer - (3)

Explanation:
If the last three digits of a number is divisible by 8, then the number is divisible by 8 (test of divisibility by 8).
Here, last three digits 58N is divisible by 8 if N = 4. (Since 584 is divisible by 8.)

For divisibility by 11. If the digits at odd and even places of a given number are equal or differ by a number divisible by 11, then the given number is divisible by 11.

Therefore, (M+9+4+4+8)-(3+0+8+5+N)=(M+5) should be divisible by 11 => when M = 6.

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Question:
How many zeros contained in 100!?

(1) 100
(2) 24
(3) 97
(4) Cannot be determined

Correct Answer - (2)

Explanation:
If the number N = 2×1.3×2.5×3 ….. pxq. It is clear that each pair of prime factors 2 and 5 will generate one zero in the number N, because 10 = 2X5.

For finding out the number of zeros in the number it is sufficient to find out how many two’s and five’s appear in the expansion of each product N.

No. of five’s = No. of two’s = Hence it is evident that there are only 24 pairs of primes 2 and 5 and therefore 100! ends in 24 zeros.

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