how many numbers between 200 and 600 are divisible by 4,5 & 6?
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Answered by
3
★★★hey friend!!!!★★★
_____________________________________________________________
⇒ here is urs answer!!
Let’s firstly drop some rules of divisibility by 4, 5 and 6:
→ by 4: The last two digits are divisible by 4
→ by 5: The last digit is either a 0 or a 5
→ by 6: The number is even (divisible by two) and its sum is divisible by three (divisible by three)
⇒ Now, if the last two digits are divisible by 4, they will not end in a 5. Therefore, we are left with the possible terminations for our numbers:
→ 00
→ 20
→ 40
→ 60
→ 80
⇒ We just need to choose the first digit so that the sum of the digits is divisible by 3. That yelds this:
⇒ 00→3or6
⇒ 40→2
⇒ 60→3
⇒ 80→5
This means that there are 6 numbers between 200 and 600 which are simultaneously divisible by 4, 5 and 6
___________________________________________________________
★★★hope this may help you!!★★★
_____________________________________________________________
⇒ here is urs answer!!
Let’s firstly drop some rules of divisibility by 4, 5 and 6:
→ by 4: The last two digits are divisible by 4
→ by 5: The last digit is either a 0 or a 5
→ by 6: The number is even (divisible by two) and its sum is divisible by three (divisible by three)
⇒ Now, if the last two digits are divisible by 4, they will not end in a 5. Therefore, we are left with the possible terminations for our numbers:
→ 00
→ 20
→ 40
→ 60
→ 80
⇒ We just need to choose the first digit so that the sum of the digits is divisible by 3. That yelds this:
⇒ 00→3or6
⇒ 40→2
⇒ 60→3
⇒ 80→5
This means that there are 6 numbers between 200 and 600 which are simultaneously divisible by 4, 5 and 6
___________________________________________________________
★★★hope this may help you!!★★★
kushanaanandp4ac8a:
We can do this by AP too
Answered by
2
Hey,sup!
As per the question,
HCF of 4,5,6 is 60.
So first term(a)= 240 (in the range which is divisible by 60.
Last term(l)= 600 ( in the range which is divisible by 60.
difference(d)= 60.
=>T(n)=a+(n-1)d.
=>600=240+(n-1)60.
=>600-240=(n-1)60.
=>360=(n-1)60.
=>n-1=360/60.
=>n-1=6.
=>n=6+1=7.
So,there are 7 numbers between 200 to 600 that are divisible by 4,5,6.
Hope it helps.
As per the question,
HCF of 4,5,6 is 60.
So first term(a)= 240 (in the range which is divisible by 60.
Last term(l)= 600 ( in the range which is divisible by 60.
difference(d)= 60.
=>T(n)=a+(n-1)d.
=>600=240+(n-1)60.
=>600-240=(n-1)60.
=>360=(n-1)60.
=>n-1=360/60.
=>n-1=6.
=>n=6+1=7.
So,there are 7 numbers between 200 to 600 that are divisible by 4,5,6.
Hope it helps.
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