Question

5 anko ki badi Se Badi sankhya gyat Kijiye jise 9 ,12 ,15 ,18 se Bhag dene per 6 ,9 ,12 ,15 shesh bache​

Answer 1

SOLUTION

TO DETERMINE

The greatest number of five digits which on dividing by 9 ,12 ,15 ,18 leaves 6 ,9 ,12 ,15 respectively as remainders.

EVALUATION

Here the given divisors are 9 ,12 ,15 ,18

Now the LCM of 9 ,12 ,15 ,18 = 180

We know that the greatest number of five digits is 99999

If we divide 99999 by 180 we get 555 as Quotient and 99 as remainder

So the greatest number of five digits which is divisible by 180 is

= 99999 - 99

= 99900

Again the divisors are 9 ,12 ,15 ,18 and the Remainders are 6 ,9 ,12 ,15

Now

$9 - 6 = 3$

$12 - 9 = 3$

$15 - 12 = 3$

$18 - 15 = 3$

So the difference between Divisor and Remainder is 3

Hence the required number is

= 99900 - 3

= 99897

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