Question

Find the greatest 4 digit number which when divided by 20, 24 and 45 leaves a remainder 10 III
case.
Find the least 5 digit number which leaves a remainder 9 in each case when divided by 12, 40 and 75.
he vontact three digit number divisible by 8 and 18. 1 930
10209
T​

Answer 1

Answer:

the LCM of these 3 numbers = 360

the highest 4 digit number divisible by 360 = 9720

the required answer is 9720 +10 = 9730

Step-by-step explanation:

hope it helped you

please follow me

mark this answer as brilliant answer

Answer 2

Answer:

Answer

To find the least 5-digit number which leaves a remainder 9 in each case when they are divided by 12,40 and 75

Let us see factors for the given numbers 12,40,75

For 12 prime factors are 12:2  

2

×3

for 40 prime factors are 40:=2  

3

×5

For 75 prime factors are 75=5  

2

×3

So, now let us find out the greatest four digit number that which is exactly divisible by given numbers

∴ the greatest four digit number divisible by given numbers =9999

So, LCM of the given numbers is LCM=2  

3

×3×5  

2

=600

So, to find out the greatest four digit divisible by given numbers ⇒ 9999−remainder

⇒ 9999−399=9600

in order to get 5 digit number exactly divisible by the given numbers , we get

9600+600=10200

but given in the question that when 5-digit number is divided by the given numbers we get a remainder 9

as it is greatest number we are finding, here we subtract remainder to 5-digit number, for least number we add the remainder from it

So, we get, 10200+9=10209

∴ 10209 is the least 5-digit number which when divided by 12,40 and 75 leaves a remainder 9.

Step-by-step explanation:

hope It's Helpful!!