Find the greatest number of 5 digits which when divided by 25, 30 and 40 leaves a
remainder 20,25, and 35 respectively.

The greatest 5 digit number which when divided by 25, 30 and 40 has a remainder of 20, 25, and 35 is 99,595

The reason why the above value arrived at is correct is as follows:

The required parameter;

To find a 5 digit number that with remainder of 20, remainder of 25 and a remainder of 35, when divided by 25, 30, and 40 respectively

Strategy;

Find the LCM of the divisor, then find the highest common multiple of the

LCM that is a 5 digit number, equate the expression for the required 5 digit

number to the highest common multiple of LCM of the divisors by adding

a value that will give a factor of the divisor as follows;

Let x represent the 5 digit number, from the question, we get;

x = 25·a + 20

x = 30·b + 25

x = 40·c + 35

x < 99,999

The 5 digit number is not a multiple of 25, 30, and 40, therefore, the number is not a multiple of the LCM of 25, 30, and 40 which is 600

The highest multiple of 600 which is a 5 digit number = 99,600

Therefore, we can write;

25·a + 20 + 5 = 30·b + 25 + 5 = 40·c + 35 + 5 = 99,600

However;

25·a + 20 + 5 = x + 5

By transitive property, we get

x + 5 = 99,600

∴ x = 99,600 - 5 = 99,595

The 5 digit number, x = 99,595

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99,595

Step-by-step explanation:

We are looking at division and remainders. What do you do when a remainder is present? You usually add it when multiplying to get the final number, but since we are doing this backwards, we have to subtract the numbers divided by the remainder.

25 - 20 = 5

30 - 25 = 5

40 - 35 = 5

Look for the LCM (Least Common Multiple) for 20, 30, and 40 :

600

Now for the equation :

n + 5 = Multiple of all numbers and LCM =

n + 5 = 166 * 600 = 99,600

n = 99,600 - 5 = 99,595

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step-by-step explanation:

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