**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 â€“ 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 â€“ 7x + 12 â€“ k perfectly (k being the number to be subtracted), the remainder is 0 when the value of x is substituted by â€“3.

=> (-3)3 + 4(-3)2 â€“ 7(-3) + 12 â€“ 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 â€¦.. 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 twoâ€™s and fiveâ€™s appear in the expansion of each product N.

No. of fiveâ€™s = No. of twoâ€™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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