QBManager

Permutation and Combination

119210 If $15^{k}$ divides 47 ! But $15^{k + 1}$ does not divide it, then $k =$

1 15

2 12

3 10

4 5

Explanation:

C $: 15^{k} divided by = 47!$
$15 = 3 \times 5$
$∵ The largest power of prime P which divides by n! is$
$[\frac{n}{p}] + [\frac{n}{P^{2}}] + [\frac{n}{P^{3}}] + \dots \dots . .$
$Therefore, the largest power 3 which divide 47!$
$[\frac{47}{3}] + [\frac{47}{3^{2}}] + [\frac{47}{3^{3}}] + [\frac{47}{3^{4}}] + \dots \dots . .$
$= 15 + 5 + 1 = 21$
$∵$ The largest power of prime $P$ which divides by $n$ ! is
Therefore, the largest power 3 which divide 47 !
$⌊ \frac{47}{3} ⌋ + ⌊ \frac{47}{3^{2}} ⌋ + ⌊ \frac{47}{3^{3}} ⌋ + [\frac{47}{3^{4}} ⌋ + \dots \dots \dots$
$= 15 + 5 + 1 = 21$
The largest power of 5 which divide 47 !
$[\frac{47}{5}] + [\frac{47}{5^{2}}] + [\frac{47}{5^{3}}] + \dots \dots \dots .$
$9 + 1 = 10$ So, the largest exponent of 15 coincide $47! = 10$

AP EAMCET-20.04.2019

Permutation and Combination

119213 The remainder obtained when $1! + 2! + 3! + \dots + 11$ ! is divided by 12 is

1 9

2 8

3 7

4 6

Explanation:

A Given that,
$1! + 2! + 3! + \dots \dots \dots 11$ !
$= (1! + 2! + 3!) + (4! + 5! + \dots \dots \dots 11!)$
$= 9 + (4! + 5! + \dots \dots \dots 11!)$
$(4! + 5! + 6! + \dots \dots .11!)$ are divisible by 12
Hence, remainder will be 9 .

WB JEE-2014

Permutation and Combination

119214 The ten's digit in
$1! + 4! + 7$ ! +10! + 12! + 13! + 15! + 16! + 17! is divisible by

1 4 !

2 3 !

3 5 !

4 7 !

Explanation:

B The given expression
$1! + 4! + 7! + 10! + 12! + 13! + 15! + 17!$
We can observe that from $10!$ onwards both units and tens place are zeros.
While calculating tens place in series they all end with $^{'} z e r o - z e r o^{'}$
$10! = 3628800$
$11! = 39916800$
$N o w,$
$1! = 1$
$4! = 24$
$7! = 5040$
$= 1! + 4! + 7!$
$= 1 + 24 + 5040 = 5065$
Tens digit is 6
So, it is divisible by $3!$

AP EAMCET-22.09.2020

Permutation and Combination

119215 How many multiples of 5 are there 10 to 95 including both 10 and 95 ?

1 17

2 18

3 16

4 19

Explanation:

B The given range of numbers
$= 10$ to 95
$= 10, 15, 20 \dots \dots \dots . . . .95$
Term of last digit
$T_{l} = a + (n - 1) d$
$95 = 10 + (n - 1) 5$
$85 = (n - 1) 5$
$(n - 1) = 17$
$n = 18$

AP EAMCET-22.09.2020

Permutation and Combination

119210 If $15^{k}$ divides 47 ! But $15^{k + 1}$ does not divide it, then $k =$

1 15

2 12

3 10

4 5

Explanation:

C $: 15^{k} divided by = 47!$
$15 = 3 \times 5$
$∵ The largest power of prime P which divides by n! is$
$[\frac{n}{p}] + [\frac{n}{P^{2}}] + [\frac{n}{P^{3}}] + \dots \dots . .$
$Therefore, the largest power 3 which divide 47!$
$[\frac{47}{3}] + [\frac{47}{3^{2}}] + [\frac{47}{3^{3}}] + [\frac{47}{3^{4}}] + \dots \dots . .$
$= 15 + 5 + 1 = 21$
$∵$ The largest power of prime $P$ which divides by $n$ ! is
Therefore, the largest power 3 which divide 47 !
$⌊ \frac{47}{3} ⌋ + ⌊ \frac{47}{3^{2}} ⌋ + ⌊ \frac{47}{3^{3}} ⌋ + [\frac{47}{3^{4}} ⌋ + \dots \dots \dots$
$= 15 + 5 + 1 = 21$
The largest power of 5 which divide 47 !
$[\frac{47}{5}] + [\frac{47}{5^{2}}] + [\frac{47}{5^{3}}] + \dots \dots \dots .$
$9 + 1 = 10$ So, the largest exponent of 15 coincide $47! = 10$

AP EAMCET-20.04.2019

Permutation and Combination

119213 The remainder obtained when $1! + 2! + 3! + \dots + 11$ ! is divided by 12 is

1 9

2 8

3 7

4 6

Explanation:

A Given that,
$1! + 2! + 3! + \dots \dots \dots 11$ !
$= (1! + 2! + 3!) + (4! + 5! + \dots \dots \dots 11!)$
$= 9 + (4! + 5! + \dots \dots \dots 11!)$
$(4! + 5! + 6! + \dots \dots .11!)$ are divisible by 12
Hence, remainder will be 9 .

WB JEE-2014

Permutation and Combination

119214 The ten's digit in
$1! + 4! + 7$ ! +10! + 12! + 13! + 15! + 16! + 17! is divisible by

1 4 !

2 3 !

3 5 !

4 7 !

Explanation:

B The given expression
$1! + 4! + 7! + 10! + 12! + 13! + 15! + 17!$
We can observe that from $10!$ onwards both units and tens place are zeros.
While calculating tens place in series they all end with $^{'} z e r o - z e r o^{'}$
$10! = 3628800$
$11! = 39916800$
$N o w,$
$1! = 1$
$4! = 24$
$7! = 5040$
$= 1! + 4! + 7!$
$= 1 + 24 + 5040 = 5065$
Tens digit is 6
So, it is divisible by $3!$

AP EAMCET-22.09.2020

Permutation and Combination

119215 How many multiples of 5 are there 10 to 95 including both 10 and 95 ?

1 17

2 18

3 16

4 19

Explanation:

B The given range of numbers
$= 10$ to 95
$= 10, 15, 20 \dots \dots \dots . . . .95$
Term of last digit
$T_{l} = a + (n - 1) d$
$95 = 10 + (n - 1) 5$
$85 = (n - 1) 5$
$(n - 1) = 17$
$n = 18$

AP EAMCET-22.09.2020

Permutation and Combination

119210 If $15^{k}$ divides 47 ! But $15^{k + 1}$ does not divide it, then $k =$

1 15

2 12

3 10

4 5

Explanation:

C $: 15^{k} divided by = 47!$
$15 = 3 \times 5$
$∵ The largest power of prime P which divides by n! is$
$[\frac{n}{p}] + [\frac{n}{P^{2}}] + [\frac{n}{P^{3}}] + \dots \dots . .$
$Therefore, the largest power 3 which divide 47!$
$[\frac{47}{3}] + [\frac{47}{3^{2}}] + [\frac{47}{3^{3}}] + [\frac{47}{3^{4}}] + \dots \dots . .$
$= 15 + 5 + 1 = 21$
$∵$ The largest power of prime $P$ which divides by $n$ ! is
Therefore, the largest power 3 which divide 47 !
$⌊ \frac{47}{3} ⌋ + ⌊ \frac{47}{3^{2}} ⌋ + ⌊ \frac{47}{3^{3}} ⌋ + [\frac{47}{3^{4}} ⌋ + \dots \dots \dots$
$= 15 + 5 + 1 = 21$
The largest power of 5 which divide 47 !
$[\frac{47}{5}] + [\frac{47}{5^{2}}] + [\frac{47}{5^{3}}] + \dots \dots \dots .$
$9 + 1 = 10$ So, the largest exponent of 15 coincide $47! = 10$

AP EAMCET-20.04.2019

Permutation and Combination

119213 The remainder obtained when $1! + 2! + 3! + \dots + 11$ ! is divided by 12 is

1 9

2 8

3 7

4 6

Explanation:

A Given that,
$1! + 2! + 3! + \dots \dots \dots 11$ !
$= (1! + 2! + 3!) + (4! + 5! + \dots \dots \dots 11!)$
$= 9 + (4! + 5! + \dots \dots \dots 11!)$
$(4! + 5! + 6! + \dots \dots .11!)$ are divisible by 12
Hence, remainder will be 9 .

WB JEE-2014

Permutation and Combination

119214 The ten's digit in
$1! + 4! + 7$ ! +10! + 12! + 13! + 15! + 16! + 17! is divisible by

1 4 !

2 3 !

3 5 !

4 7 !

Explanation:

B The given expression
$1! + 4! + 7! + 10! + 12! + 13! + 15! + 17!$
We can observe that from $10!$ onwards both units and tens place are zeros.
While calculating tens place in series they all end with $^{'} z e r o - z e r o^{'}$
$10! = 3628800$
$11! = 39916800$
$N o w,$
$1! = 1$
$4! = 24$
$7! = 5040$
$= 1! + 4! + 7!$
$= 1 + 24 + 5040 = 5065$
Tens digit is 6
So, it is divisible by $3!$

AP EAMCET-22.09.2020

Permutation and Combination

119215 How many multiples of 5 are there 10 to 95 including both 10 and 95 ?

1 17

2 18

3 16

4 19

Explanation:

B The given range of numbers
$= 10$ to 95
$= 10, 15, 20 \dots \dots \dots . . . .95$
Term of last digit
$T_{l} = a + (n - 1) d$
$95 = 10 + (n - 1) 5$
$85 = (n - 1) 5$
$(n - 1) = 17$
$n = 18$

AP EAMCET-22.09.2020

Permutation and Combination

119210 If $15^{k}$ divides 47 ! But $15^{k + 1}$ does not divide it, then $k =$

1 15

2 12

3 10

4 5

Explanation:

C $: 15^{k} divided by = 47!$
$15 = 3 \times 5$
$∵ The largest power of prime P which divides by n! is$
$[\frac{n}{p}] + [\frac{n}{P^{2}}] + [\frac{n}{P^{3}}] + \dots \dots . .$
$Therefore, the largest power 3 which divide 47!$
$[\frac{47}{3}] + [\frac{47}{3^{2}}] + [\frac{47}{3^{3}}] + [\frac{47}{3^{4}}] + \dots \dots . .$
$= 15 + 5 + 1 = 21$
$∵$ The largest power of prime $P$ which divides by $n$ ! is
Therefore, the largest power 3 which divide 47 !
$⌊ \frac{47}{3} ⌋ + ⌊ \frac{47}{3^{2}} ⌋ + ⌊ \frac{47}{3^{3}} ⌋ + [\frac{47}{3^{4}} ⌋ + \dots \dots \dots$
$= 15 + 5 + 1 = 21$
The largest power of 5 which divide 47 !
$[\frac{47}{5}] + [\frac{47}{5^{2}}] + [\frac{47}{5^{3}}] + \dots \dots \dots .$
$9 + 1 = 10$ So, the largest exponent of 15 coincide $47! = 10$

AP EAMCET-20.04.2019

Permutation and Combination

119213 The remainder obtained when $1! + 2! + 3! + \dots + 11$ ! is divided by 12 is

1 9

2 8

3 7

4 6

Explanation:

A Given that,
$1! + 2! + 3! + \dots \dots \dots 11$ !
$= (1! + 2! + 3!) + (4! + 5! + \dots \dots \dots 11!)$
$= 9 + (4! + 5! + \dots \dots \dots 11!)$
$(4! + 5! + 6! + \dots \dots .11!)$ are divisible by 12
Hence, remainder will be 9 .

WB JEE-2014

Permutation and Combination

119214 The ten's digit in
$1! + 4! + 7$ ! +10! + 12! + 13! + 15! + 16! + 17! is divisible by

1 4 !

2 3 !

3 5 !

4 7 !

Explanation:

B The given expression
$1! + 4! + 7! + 10! + 12! + 13! + 15! + 17!$
We can observe that from $10!$ onwards both units and tens place are zeros.
While calculating tens place in series they all end with $^{'} z e r o - z e r o^{'}$
$10! = 3628800$
$11! = 39916800$
$N o w,$
$1! = 1$
$4! = 24$
$7! = 5040$
$= 1! + 4! + 7!$
$= 1 + 24 + 5040 = 5065$
Tens digit is 6
So, it is divisible by $3!$

AP EAMCET-22.09.2020

Permutation and Combination

119215 How many multiples of 5 are there 10 to 95 including both 10 and 95 ?

1 17

2 18

3 16

4 19

Explanation:

B The given range of numbers
$= 10$ to 95
$= 10, 15, 20 \dots \dots \dots . . . .95$
Term of last digit
$T_{l} = a + (n - 1) d$
$95 = 10 + (n - 1) 5$
$85 = (n - 1) 5$
$(n - 1) = 17$
$n = 18$

AP EAMCET-22.09.2020

Question Bank Series

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Question Bank Series

Explanation:

Explanation:

Explanation:

Explanation: