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QUESTION: 1

Find the remainder when 73 + 75 + 78 + 57 + 197 is divided by 34.

Solution:

The remainder would be given by: (5 + 7 + 10 + 23 + 27)/34 = 72/34

-> remainder = 4. Option (b) is correct.

QUESTION: 2

Find the remainder when 51^{203 }is divided by 7.

Solution:

51^{203}/7 -> 2^{203}/7 = (2^{3})^{67} X 22/7 = 8^{67} X 4/7 -> remainder = 4. Option (a) is correct

QUESTION: 3

Find the remainder when 21^{875 }is divided by 17.

Solution:

21^{875}/17 -> 4^{875}/17 = (4^{4})^{n} X 4^{3}/17 = 256^{n} X 64/17 -> 1^{n} X 13/17 -> remainder =13. Option (b) is correct.

QUESTION: 4

Find the number of consecutive zeroes at the end of:

57 X 60 X 30 X 15625 X 4096 X 625 X 875 X 975

Solution:

The given expression has fifteen 2’s and seventeen 5’s. The number of zeroes would be 15 as the number of 2’s is lower in this case. Option (d) is correct.

QUESTION: 5

Find the number of consecutive zeroes at the end of the following numbers. - 100! x 200!

Solution:

The number of zeroes would depend on the number of 5’s in the value of the factorial.

100! would end in 20 + 4 = 24 zeroes

200! Would end in 40 + 8 + 1 = 49 zeroes.

When you multiply the two numbers (one with 24 zeroes and the other with 49 zeroes at it’s end), the resultant total would end in 24 + 49 = 73 zeroes. Option (b) is correct.

QUESTION: 6

Find the number of consecutive zeroes at the end of the following numbers. 72!

Solution:

The number of zeroes would depend on the number

of 5’s in the value of the factorial. 72! -> 14 + 2 = 16. Option (d) is correct.

QUESTION: 7

Find the maximum value of n such that 50! is perfectly divisible by 2520".

Solution:

2520=7 X 3^{2} X 2^{3} X 5.

The value of n would be given by the value of the number of 7s in 50! This Value Is Equal To[50/7] + [50/49] = 7 + 1 = 8 Option (b) is correct.

QUESTION: 8

Find the maximum value of n such that 570 X 60 X 30 X 90 X 100 X 500 X 700 X 343 X 720 X 81 is perfectly divisible by 30^{n}.

Solution:

Checking for the number of 2’s, 3’s and 5’s in the given expression you can see that the minimum is for the number of 3’s (there are 11 of them while there are 12 5’s and more than 11 2’s) Hence, option (b) is correct.

QUESTION: 9

Find the maximum value of n such that 77! is per-fectly divisible by 720^{n}.

Solution:

720 = 2^{4} X 5^{1} X 3^{2}

In 77! there would be 38 + 19 + 9 + 4 + 2 + 1 = 73 twos Æ hence [73/4] = 18 2^{4s}

In 77! there would be 25 + 8 + 2 = 35 threes -> hence [35/2] = 17 3^{2s}

In 77! there would be 15 + 3 = 18 fives

Since 17 is the least of these values, option (c) is correct.

QUESTION: 10

Find the number of consecutive zeroes at the end of the following numbers. 100! + 200!

Solution:

The number of zeroes would depend on the number of 5’s in the value of the factorial. 100! wouldendin 20 + 4 = 24zeroes 200!Wouldendin 40 + 8 + 1 = 49zeroes.

When you add the two numbers (one with 24 zeroes and the other with 49 zeroes at it’s end), the resultant total would end in 24 zeroes. Option (b) is correct

### Divisibility Test for 5

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### Divisibility Test for 10

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### Divisibility Test for 3, 6, 9

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### Competition Level Test: Solutions

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