Question

Find the least five-digit number which when divided by 20, 40, 75 leaves remainder 9 in each case. .....
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Answer 1

Answer:

10209

Step-by-step explanation:

20 = 2 * 2 * 5

40 = 2 * 2 * 2 * 5

75 = 3 * 5 * 5

LCM = 2 * 2 * 2 * 5 * 5 * 3 = 600

Least 5 digit number = 10000.

600)10000( 16

- 9600

400

So, the least 5 digit number divisible by these numbers is 10000 + 600 - 400 = 10200

Therefore required number is 10200 + 9 = 10209

Answer 2

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=

${5}^{3}$

×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.

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Sorry if it is wrong