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Divisibility

Divisibility

MODULES

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Divisibility - 2, 5, 10
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Divisibility - 3, 9, 11
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7, 13 and Composite
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Divisibility of Factorial
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Last Digit
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Last 2 Digits
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Past Questions

CONCEPTS & CHEATSHEET

Concept Revision Video

SPEED CONCEPTS

Divisibility 1
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Divisibility 2
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PRACTICE

Divisibility : Level 1
Divisibility : Level 2
Divisibility : Level 3
ALL MODULES

CAT 2025 Lesson : Divisibility - Divisibility of Factorial

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3. Factorials

Factorial is represented by an exclamation mark (!).

'n!', read as 'n factorial', is the product of all natural numbers less than or equal to n.
∴
n!=1×2×... ×nn! = 1 \times 2 \times ... \space \times nn!=1×2×... ×n

Exception: Although
000 is not a natural number, 0!=1\bm{0! = 1}0!=1.

3.1 Power of a prime number in a factorial

Let's look at this with an example. The highest power of 555 that can divide 100!100!100!, is the power of 555 when 100!100!100! is prime factorised.

100!=1×2×3×... ×100100! = 1 \times 2 \times 3 \times ... \space \times 100100!=1×2×3×... ×100

100!100!100! contains 20\bm{20}20 multiples of 5\bm{5}5 ⇒ 5×1, 5×2, ..., 5×20 5 \times 1, \space 5 \times 2, \space ..., \space 5 \times 205×1, 5×2, ..., 5×20

In these, there are
4\bm{4}4 multiples of 52\bm{5^2}52 ⇒ 25×1, 25×2, 25×3, 25×4 25 \times 1, \space 25 \times 2, \space 25 \times 3, \space 25 \times 425×1, 25×2, 25×3, 25×4

As the multiples of
525^252 were already counted as multiples of 555, we simply add them one more time.
∴ The highest power of
555 which divides 100!=20+4=24100! = 20 + 4 = \bm{24}100!=20+4=24

3.1.1 Successive Division

An easier way to find this is through successive division. The highest power of a prime number x, that divides
n!\bm{n!}n! is the sum of quotients when n is successively divided by xxx (as shown below).



∴ The highest power of
555 which divides 100!=20+4+0=24100! = 20 + 4 + 0 = \bm{24}100!=20+4+0=24

Example 9

What is the highest power of 777 that divides 1000!1000!1000!?

Solution



∴ The highest power of 777 which divides 1000!=142+20+2=1641000! = 142 + 20 + 2 = \bm{164}1000!=142+20+2=164

Answer:
164164164

3.2 Factorials and division of composite numbers

Let
x=apbqcrx = a^{p}b^{q}c^{r}x=apbqcr, where aaa, bbb and ccc are prime numbers. To find the largest power of xxx that would perfectly divide a given factorial, say n!n!n!.

Step 1: Prime factorise
xxx and write it as x=apbqcrx = a^{p}b^{q}c^{r}x=apbqcr
Step 2: Find the highest power of
aaa, bbb and ccc that can perfectly divide n!n!n!
Step 3: Divide the highest power of
aaa, bbb and ccc from Step 222 by ppp, qqq and rrr respectively and note the quotients.
Step 4: The lowest quotient from Step
333 is the answer.

Example 10

What is the highest power of 242424 that divides 90!90!90!?

Solution

Step 1: 24=23×3124 = 2^{3} \times 3 ^{1}24=23×31

Step 2:



Step 3:

The highest power of
23=Quot(863)=282^{3} = \text{Quot} \left(\dfrac{86}{3} \right) = 2823=Quot(386​)=28

The highest power of
31=Quot(441)=443^{1} = \text{Quot} \left(\dfrac{44}{1} \right) = 4431=Quot(144​)=44

Step 4: Lowest quotient in step 3 is
282828.

∴ The highest power of
242424 which divides 90!=2890! = \bm{28}90!=28

Answer:
282828

3.3 Number of zeroes

The number of zeroes at the end of a factorial will be the highest power of
555 that divides the factorial. This is explained in the example below.

Example 11

When 200!200!200! is written as a natural number, how many consecutive zeroes are there in the extreme right of the number?

Solution

We get nnn zeroes at the end of a number if it is a multiple of 10n10^{n}10n.
∴ We need to find the highest power of 10 that divides
200!200!200!.

10=21×5110 = 2^{1} \times 5^{1}10=21×51
As the power of both the prime factors is the same (which is
111), the larger prime will be occurring fewer number of times in a given factorial.

∴ Highest power of
222 which divides 200!200!200! > Highest power of 555 which divides 200!200!200!

As we are concerned with the smaller of these powers, the highest power of
555 which divides 200!200!200! will be the highest power of 101010 which divides 200!200!200!.



Number of zeroes at the end of
200! =200! \space =200! = The highest power of 555 in 200!200!200!
=40+8+1=49= 40 + 8 + 1 = 49=40+8+1=49

Answer:
494949


Example 12

If n!n!n! has 191919 zeroes right before the first non-zero digit, then which of the following could be the value of nnn?

(1)
757575            (2) 838383            (3) 878787            (4) 909090           

Solution

As explained above, to find the number of zeroes at the end, we need to find the highest power of 555 that divides n!n!n!

We start with the smallest number in the options.



Highest power of
555 that divides 75!=15+3=18\bm{75!} = 15 + 3 = \bm{18}75!=15+3=18

83!83!83! has one more 555 multiple (which is 808080) than 75!75!75!.

Highest power of
555 that divides 83!=18+1=19\bm{83!} = 18 + 1 = \bm{19}83!=18+1=19

∴
83!\bm{83!}83! will have 19 zeroes\bm{19 \space zeroes}19 zeroes at the end.

Answer:
838383
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