Question

Find the greatest 6 - digit number which on dividing by 42,45,48,56and60 leaving remainder in 12 each case​

Answer 1

Answer:

The LCM of 3,12,18,24,30,36 is 360

We have the greatest five digit number : 99999

When 99999 is divided by 360, leaves the remainder 279

So, 99999−279=99720

Now here, we can see that

3−2=1, 12−11=1, 18−17=1, 24−23=1, 30−29=1, 36−35=1

The difference of divisors and their corresponding remainders is 1.

So, the required five digit number will be 1 less than $99720$$.

The required number is 99720−1=99719

Step-by-step explanation:

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Answer 2

Answer:

Prime factor of 20,30,35and45

20=2

2

×5

30=2×3×5

35=5×7

45=3

2

×5

LCM=product of each prime factor of highest power

LCM=2

2

×3²×5×7=1260

Greatest four digit number=9999

greatest four digit number divisible by all given numbers=9999-remainder when 9999 is divided by LCM of given numbers

greatest four digit number divisible by given numbers=9999-1260=8739

Given that,

required number when divided by 20 , 30, 35,45 leaves remainder 12.

Therefore,required number=8739+12=8751

Hence greatest four digit number when divided by 20,30,35 and 45 is 8751.

Step-by-step explanation:

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