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Sanjay Sharma
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in Probability
Dec 27, 2021

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**If a number is divisible by 4 and 5, then it is divisible by 20**** (since 4 and 5 are relatively prime)…. **

**A number is divisible by 20 if it ends in 20,40,60,80 or 00....**

**There exist two cases...**

**Numbers ending in 20,40,60,80, and numbers ending in 00…**

** I. Number of three-digit numbers divisible by 20 ending in 20,40,60,80 =4×9 =36...**

**Possible permutations of each of these three-digit number, including the number itself **

**=4 (e.g. 520→502,205,250) ...**

**Therefore, numbers in this case = 36×4 =144 ...**

**However, when both non-zero digits are even, the permutations are repeated. **

**(e.g. 820→802,208,280 and 280→802,208,820) **

**Such numbers =24×4=8 …**

**Subtract these cases. ⇒ total numbers in this case =144−8 =136 ..**

**For those numbers ending in 00, no other permutations are possible.**

**Hence, numbers in this case =9 ..**

**Total cases **

**=136+9=145 ... **

**Total 3-digit numbers **

**=9×10×10=900 .... **

**=( 145/900 ) = ( 29 / 180 ) .…**

avoid copy paste while answering,

case 1 with two “00” total 9 such numbers are possible (100 ,200..900)

case 2 with one 0 at least one digit should be even so 3 sub cases

a)both digit are even and distinct (4x3)x4 =48 such numbers (204,240,420,402 4 permutation)

b) both digits are even and same (4x1)x2=8 (e.g 220 and 202 only 2 permutation)

c) one digit even other odd (5x4)x4 =80 (104,140,410,401 4 permutation)

total 9+48+8+80=145 such numbers

prob =145/900 =29/180

case 1 with two “00” total 9 such numbers are possible (100 ,200..900)

case 2 with one 0 at least one digit should be even so 3 sub cases

a)both digit are even and distinct (4x3)x4 =48 such numbers (204,240,420,402 4 permutation)

b) both digits are even and same (4x1)x2=8 (e.g 220 and 202 only 2 permutation)

c) one digit even other odd (5x4)x4 =80 (104,140,410,401 4 permutation)

total 9+48+8+80=145 such numbers

prob =145/900 =29/180

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