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- Index of Greatest power
a. 4
b. 5
c. 6
d. 1
Answer: B
Explanation:
Concept:Index of greatest power (IGP) of a number is the maximum power of that number that divides the original expression. The first step in finding IGP is to write the given number in prime factorization format.
Prime factorization of $1600 = {2^\boxed6} \times {5^2}$
IGP of 2 in 1600 is 6.
Concept:Index of greatest power (IGP) of a number is the maximum power of that number that divides the original expression. The first step in finding IGP is to write the given number in prime factorization format.
Prime factorization of $1600 = {2^\boxed6} \times {5^2}$
IGP of 2 in 1600 is 6.
2. Find the index of greatest power of 4 in 1600.
a. 3
b. 5
c. 6
d. 9
Answer: A
Explanation:
$1600 = {2^6} \times {5^2}$
To get the index of 4, we need to pull out 4's from the above expression.
$\therefore$ $1600 = {\left( {{2^2}} \right)^3} \times {5^2}$ $ = {\text{ }}{4^3} \times {5^2}$
$\therefore$ Index of 4 is 3.
$1600 = {2^6} \times {5^2}$
To get the index of 4, we need to pull out 4's from the above expression.
$\therefore$ $1600 = {\left( {{2^2}} \right)^3} \times {5^2}$ $ = {\text{ }}{4^3} \times {5^2}$
$\therefore$ Index of 4 is 3.
3. Find the maximum power of 5 in 60!
a. 10
b. 12
c. 13
d. 14
Answer: D
Explanation:
Writing $60!$ in prime factorization format is difficult. So we only focus on the power of $5$ when $60!$ is written in prime factorization format.
$60! = 1 \times 2 \times 3 ...\times\boxed5$. . .$\times\boxed{10}...\times\boxed{25}$. . .$\times\boxed{50}$. . .$\times\boxed{60}$
$60! = 1 \times 2 \times 3 ...\times\boxed5$. . .$\times\boxed{2 \times 5}$. . .$\times\boxed{5^2}$. . .$\times\boxed{2\times5^2}$. . .$\times\boxed{5 \times 12}$
Every fifth number is a multiple of 5.
So there must be 60/5 = 12 fives are available.
In addition to this $25 = 5^2 $ and $50 = 2 \times 5^2$ contribute another two 5's. So total number is 12 + 2 = 14
$\left[ {\displaystyle\frac{{60}}{5}} \right] + \left[ {\displaystyle\frac{{60}}{{{5^2}}}} \right] = 12 + 2 = 14$
Here [ ] Indicates greatest integer function.
Shortcut:
Divide 60 by 5 and write quotient. Omit any remainders. Again divide the quotient by 5. Omit any remainder. Follow the procedure, till the quotient not divisible further. Add all the numbers below the given number. The result is the answer.
Writing $60!$ in prime factorization format is difficult. So we only focus on the power of $5$ when $60!$ is written in prime factorization format.
$60! = 1 \times 2 \times 3 ...\times\boxed5$. . .$\times\boxed{10}...\times\boxed{25}$. . .$\times\boxed{50}$. . .$\times\boxed{60}$
$60! = 1 \times 2 \times 3 ...\times\boxed5$. . .$\times\boxed{2 \times 5}$. . .$\times\boxed{5^2}$. . .$\times\boxed{2\times5^2}$. . .$\times\boxed{5 \times 12}$
Every fifth number is a multiple of 5.
So there must be 60/5 = 12 fives are available.
In addition to this $25 = 5^2 $ and $50 = 2 \times 5^2$ contribute another two 5's. So total number is 12 + 2 = 14
$\left[ {\displaystyle\frac{{60}}{5}} \right] + \left[ {\displaystyle\frac{{60}}{{{5^2}}}} \right] = 12 + 2 = 14$
Here [ ] Indicates greatest integer function.
Shortcut:
Divide 60 by 5 and write quotient. Omit any remainders. Again divide the quotient by 5. Omit any remainder. Follow the procedure, till the quotient not divisible further. Add all the numbers below the given number. The result is the answer.
4. Find the maximum power of 15 in 100!
a. 6
b. 25
c. 27
d. 24
Answer: D
Explanation:
15 is a composite number. $15 = 3 \times 5$. So we find the maximum power of 3 and 5 in the above expression.
Important note: The maximum power of 5 in the above expression is less than the maximum power of 3 as 5 is bigger number.
From the above diagram, there are 24 fives and 48 threes available. The number of 15's can be formed = 24.
( Shortcut: To remember easily, think like, there are 24 women, and 48 men available. How many legal marriages possible? answer is 24! )
Theoretically, \(100! = {3^{48}} \times {5^{24}} \times ...\) = \({3^{24}} \times \left( {{3^{24}} \times {5^{24}}} \right) \times ...\) = \({3^{24}} \times \left( {{{15}^{24}}} \right) \times ...\)
So maximum power of 15 is 24
15 is a composite number. $15 = 3 \times 5$. So we find the maximum power of 3 and 5 in the above expression.
Important note: The maximum power of 5 in the above expression is less than the maximum power of 3 as 5 is bigger number.
( Shortcut: To remember easily, think like, there are 24 women, and 48 men available. How many legal marriages possible? answer is 24! )
Theoretically, \(100! = {3^{48}} \times {5^{24}} \times ...\) = \({3^{24}} \times \left( {{3^{24}} \times {5^{24}}} \right) \times ...\) = \({3^{24}} \times \left( {{{15}^{24}}} \right) \times ...\)
So maximum power of 15 is 24
5. Find the highest power of 12 that divide 49!.
a. 4
b. 22
c. 23
d. 26
Answer: B
Explanation:
12 is not a prime number. 12 = ${2^2} \times 3$
Now we find the maximum power of 2 in 49! and maximum power of 3 in 49! and find how many 12's can be formed.
So maximum power of 2 in 49! is 46. Therefore maximum power of ${2^2}$ in 49! is 23.
Also, maximum power of 3 in 49! = 22
\( \Rightarrow 49! = {\left( {{2^2}} \right)^{23}} \times {3^{22}} \times ...\) = \({\left( {{2^2} \times 3} \right)^{22}} \times {2^2} \times ...\)
So 22 is the maximum power of 12 that divides 49! exactly.
12 is not a prime number. 12 = ${2^2} \times 3$
Now we find the maximum power of 2 in 49! and maximum power of 3 in 49! and find how many 12's can be formed.
Also, maximum power of 3 in 49! = 22
\( \Rightarrow 49! = {\left( {{2^2}} \right)^{23}} \times {3^{22}} \times ...\) = \({\left( {{2^2} \times 3} \right)^{22}} \times {2^2} \times ...\)
So 22 is the maximum power of 12 that divides 49! exactly.
6. How many zero's are there at the end of 100!
a. 10
b. 11
c. 22
d. 24
Answer: C
Explanation:
A zero can be formed by the multiplication of 5 and 2. Since 100! contains more 2's than 5's, we can find the maximum power of 5 contained in 100!
For your understanding:
\( \Rightarrow 100! = {2^{97}} \times {5^{24}} \times ...\)
As there are only 24 fives are available, there are 24 zero's at the end of 100!
A zero can be formed by the multiplication of 5 and 2. Since 100! contains more 2's than 5's, we can find the maximum power of 5 contained in 100!
For your understanding:
As there are only 24 fives are available, there are 24 zero's at the end of 100!
Number System: BasicsNumber System: HCF and LCMNumber System: Factors and CoprimesNumber System: Divisbility Rules Number System: Power of a number in a Factorial Number System: Units digit of an expression Number System: Last two digits of an expressionNumber System: Base SystemNumber System: Last non zero digit of a factorial (LNZ)Number System: Last two non zero digits of a factorialNumber System: High level Exercise
