119210
If \(15^{\mathrm{k}}\) divides 47 ! But \(15^{\mathrm{k}+1}\) does not divide it, then \(\mathrm{k}=\)
1 15
2 12
3 10
4 5
Explanation:
C \(: 15^{\mathrm{k}} \text { divided by }=47 \text { ! }\) \(15=3 \times 5\) \(\because \text { The largest power of prime } \mathrm{P} \text { which divides by } \mathrm{n} ! \text { is }\) \({\left[\frac{\mathrm{n}}{\mathrm{p}}\right]+\left[\frac{\mathrm{n}}{\mathrm{P}^2}\right]+\left[\frac{\mathrm{n}}{\mathrm{P}^3}\right]+\ldots \ldots . .}\) \(\text { Therefore, the largest power } 3 \text { which divide } 47 \text { ! }\) \({\left[\frac{47}{3}\right]+\left[\frac{47}{3^2}\right]+\left[\frac{47}{3^3}\right]+\left[\frac{47}{3^4}\right]+\ldots \ldots . .}\) \(=15+5+1=21\) \(\because\) The largest power of prime \(\mathrm{P}\) which divides by \(\mathrm{n}\) ! is Therefore, the largest power 3 which divide 47 ! \(\left\lfloor\frac{47}{3}\right\rfloor+\left\lfloor\frac{47}{3^2}\right\rfloor+\left\lfloor\frac{47}{3^3}\right\rfloor+\left[\frac{47}{3^4}\right\rfloor+\ldots \ldots \ldots\) \(=15+5+1=21\) The largest power of 5 which divide 47 ! \({\left[\frac{47}{5}\right]+\left[\frac{47}{5^2}\right]+\left[\frac{47}{5^3}\right]+\ldots \ldots \ldots .}\) \(9+1=10\)So, the largest exponent of 15 coincide \(47 !=10\)
AP EAMCET-20.04.2019
Permutation and Combination
119213
The remainder obtained when \(1 !+2 !+3 !+\ldots+11\) ! is divided by 12 is
1 9
2 8
3 7
4 6
Explanation:
A Given that, \(1 !+2 !+3 !+\ldots \ldots \ldots 11\) ! \(=(1 !+2 !+3 !)+(4 !+5 !+\ldots \ldots \ldots 11 !)\) \(=9+(4 !+5 !+\ldots \ldots \ldots 11 !)\) \((4 !+5 !+6 !+\ldots \ldots .11 !)\) 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 \('zero-zero'\) \(10 !=3628800\) \(11 !=39916800\) \(Now,\) \(1 !=1\) \(4 !=24\) \(7 !=5040\) \(=1 !+4 !+7 !\) \(=1+24+5040=5065\) 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 \(=10,15,20 \ldots \ldots \ldots . . . .95\) Term of last digit \(\mathrm{T}_l=\mathrm{a}+(\mathrm{n}-1) \mathrm{d}\) \(95=10+(\mathrm{n}-1) 5\) \(85=(\mathrm{n}-1) 5\) \((\mathrm{n}-1)=17\) \(\mathrm{n}=18\)
119210
If \(15^{\mathrm{k}}\) divides 47 ! But \(15^{\mathrm{k}+1}\) does not divide it, then \(\mathrm{k}=\)
1 15
2 12
3 10
4 5
Explanation:
C \(: 15^{\mathrm{k}} \text { divided by }=47 \text { ! }\) \(15=3 \times 5\) \(\because \text { The largest power of prime } \mathrm{P} \text { which divides by } \mathrm{n} ! \text { is }\) \({\left[\frac{\mathrm{n}}{\mathrm{p}}\right]+\left[\frac{\mathrm{n}}{\mathrm{P}^2}\right]+\left[\frac{\mathrm{n}}{\mathrm{P}^3}\right]+\ldots \ldots . .}\) \(\text { Therefore, the largest power } 3 \text { which divide } 47 \text { ! }\) \({\left[\frac{47}{3}\right]+\left[\frac{47}{3^2}\right]+\left[\frac{47}{3^3}\right]+\left[\frac{47}{3^4}\right]+\ldots \ldots . .}\) \(=15+5+1=21\) \(\because\) The largest power of prime \(\mathrm{P}\) which divides by \(\mathrm{n}\) ! is Therefore, the largest power 3 which divide 47 ! \(\left\lfloor\frac{47}{3}\right\rfloor+\left\lfloor\frac{47}{3^2}\right\rfloor+\left\lfloor\frac{47}{3^3}\right\rfloor+\left[\frac{47}{3^4}\right\rfloor+\ldots \ldots \ldots\) \(=15+5+1=21\) The largest power of 5 which divide 47 ! \({\left[\frac{47}{5}\right]+\left[\frac{47}{5^2}\right]+\left[\frac{47}{5^3}\right]+\ldots \ldots \ldots .}\) \(9+1=10\)So, the largest exponent of 15 coincide \(47 !=10\)
AP EAMCET-20.04.2019
Permutation and Combination
119213
The remainder obtained when \(1 !+2 !+3 !+\ldots+11\) ! is divided by 12 is
1 9
2 8
3 7
4 6
Explanation:
A Given that, \(1 !+2 !+3 !+\ldots \ldots \ldots 11\) ! \(=(1 !+2 !+3 !)+(4 !+5 !+\ldots \ldots \ldots 11 !)\) \(=9+(4 !+5 !+\ldots \ldots \ldots 11 !)\) \((4 !+5 !+6 !+\ldots \ldots .11 !)\) 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 \('zero-zero'\) \(10 !=3628800\) \(11 !=39916800\) \(Now,\) \(1 !=1\) \(4 !=24\) \(7 !=5040\) \(=1 !+4 !+7 !\) \(=1+24+5040=5065\) 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 \(=10,15,20 \ldots \ldots \ldots . . . .95\) Term of last digit \(\mathrm{T}_l=\mathrm{a}+(\mathrm{n}-1) \mathrm{d}\) \(95=10+(\mathrm{n}-1) 5\) \(85=(\mathrm{n}-1) 5\) \((\mathrm{n}-1)=17\) \(\mathrm{n}=18\)
119210
If \(15^{\mathrm{k}}\) divides 47 ! But \(15^{\mathrm{k}+1}\) does not divide it, then \(\mathrm{k}=\)
1 15
2 12
3 10
4 5
Explanation:
C \(: 15^{\mathrm{k}} \text { divided by }=47 \text { ! }\) \(15=3 \times 5\) \(\because \text { The largest power of prime } \mathrm{P} \text { which divides by } \mathrm{n} ! \text { is }\) \({\left[\frac{\mathrm{n}}{\mathrm{p}}\right]+\left[\frac{\mathrm{n}}{\mathrm{P}^2}\right]+\left[\frac{\mathrm{n}}{\mathrm{P}^3}\right]+\ldots \ldots . .}\) \(\text { Therefore, the largest power } 3 \text { which divide } 47 \text { ! }\) \({\left[\frac{47}{3}\right]+\left[\frac{47}{3^2}\right]+\left[\frac{47}{3^3}\right]+\left[\frac{47}{3^4}\right]+\ldots \ldots . .}\) \(=15+5+1=21\) \(\because\) The largest power of prime \(\mathrm{P}\) which divides by \(\mathrm{n}\) ! is Therefore, the largest power 3 which divide 47 ! \(\left\lfloor\frac{47}{3}\right\rfloor+\left\lfloor\frac{47}{3^2}\right\rfloor+\left\lfloor\frac{47}{3^3}\right\rfloor+\left[\frac{47}{3^4}\right\rfloor+\ldots \ldots \ldots\) \(=15+5+1=21\) The largest power of 5 which divide 47 ! \({\left[\frac{47}{5}\right]+\left[\frac{47}{5^2}\right]+\left[\frac{47}{5^3}\right]+\ldots \ldots \ldots .}\) \(9+1=10\)So, the largest exponent of 15 coincide \(47 !=10\)
AP EAMCET-20.04.2019
Permutation and Combination
119213
The remainder obtained when \(1 !+2 !+3 !+\ldots+11\) ! is divided by 12 is
1 9
2 8
3 7
4 6
Explanation:
A Given that, \(1 !+2 !+3 !+\ldots \ldots \ldots 11\) ! \(=(1 !+2 !+3 !)+(4 !+5 !+\ldots \ldots \ldots 11 !)\) \(=9+(4 !+5 !+\ldots \ldots \ldots 11 !)\) \((4 !+5 !+6 !+\ldots \ldots .11 !)\) 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 \('zero-zero'\) \(10 !=3628800\) \(11 !=39916800\) \(Now,\) \(1 !=1\) \(4 !=24\) \(7 !=5040\) \(=1 !+4 !+7 !\) \(=1+24+5040=5065\) 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 \(=10,15,20 \ldots \ldots \ldots . . . .95\) Term of last digit \(\mathrm{T}_l=\mathrm{a}+(\mathrm{n}-1) \mathrm{d}\) \(95=10+(\mathrm{n}-1) 5\) \(85=(\mathrm{n}-1) 5\) \((\mathrm{n}-1)=17\) \(\mathrm{n}=18\)
119210
If \(15^{\mathrm{k}}\) divides 47 ! But \(15^{\mathrm{k}+1}\) does not divide it, then \(\mathrm{k}=\)
1 15
2 12
3 10
4 5
Explanation:
C \(: 15^{\mathrm{k}} \text { divided by }=47 \text { ! }\) \(15=3 \times 5\) \(\because \text { The largest power of prime } \mathrm{P} \text { which divides by } \mathrm{n} ! \text { is }\) \({\left[\frac{\mathrm{n}}{\mathrm{p}}\right]+\left[\frac{\mathrm{n}}{\mathrm{P}^2}\right]+\left[\frac{\mathrm{n}}{\mathrm{P}^3}\right]+\ldots \ldots . .}\) \(\text { Therefore, the largest power } 3 \text { which divide } 47 \text { ! }\) \({\left[\frac{47}{3}\right]+\left[\frac{47}{3^2}\right]+\left[\frac{47}{3^3}\right]+\left[\frac{47}{3^4}\right]+\ldots \ldots . .}\) \(=15+5+1=21\) \(\because\) The largest power of prime \(\mathrm{P}\) which divides by \(\mathrm{n}\) ! is Therefore, the largest power 3 which divide 47 ! \(\left\lfloor\frac{47}{3}\right\rfloor+\left\lfloor\frac{47}{3^2}\right\rfloor+\left\lfloor\frac{47}{3^3}\right\rfloor+\left[\frac{47}{3^4}\right\rfloor+\ldots \ldots \ldots\) \(=15+5+1=21\) The largest power of 5 which divide 47 ! \({\left[\frac{47}{5}\right]+\left[\frac{47}{5^2}\right]+\left[\frac{47}{5^3}\right]+\ldots \ldots \ldots .}\) \(9+1=10\)So, the largest exponent of 15 coincide \(47 !=10\)
AP EAMCET-20.04.2019
Permutation and Combination
119213
The remainder obtained when \(1 !+2 !+3 !+\ldots+11\) ! is divided by 12 is
1 9
2 8
3 7
4 6
Explanation:
A Given that, \(1 !+2 !+3 !+\ldots \ldots \ldots 11\) ! \(=(1 !+2 !+3 !)+(4 !+5 !+\ldots \ldots \ldots 11 !)\) \(=9+(4 !+5 !+\ldots \ldots \ldots 11 !)\) \((4 !+5 !+6 !+\ldots \ldots .11 !)\) 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 \('zero-zero'\) \(10 !=3628800\) \(11 !=39916800\) \(Now,\) \(1 !=1\) \(4 !=24\) \(7 !=5040\) \(=1 !+4 !+7 !\) \(=1+24+5040=5065\) 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 \(=10,15,20 \ldots \ldots \ldots . . . .95\) Term of last digit \(\mathrm{T}_l=\mathrm{a}+(\mathrm{n}-1) \mathrm{d}\) \(95=10+(\mathrm{n}-1) 5\) \(85=(\mathrm{n}-1) 5\) \((\mathrm{n}-1)=17\) \(\mathrm{n}=18\)
