find the greatest number of 5 digite exactly divisible by 12,15and 36 ...
find the greatest number of 5 digite exactly divisible by 12,15and 36 ...
Introduction
To find the greatest number of 5 digits exactly divisible by 12, 15, and 36, we need to use the concept of Euclid's Division Lemma.
Euclid's Division Lemma
Euclid's Division Lemma states that "given positive integers a and b, there exist unique integers q and r such that a = bq + r, where 0 ≤ r < />
Steps to find the greatest number of 5 digits exactly divisible by 12, 15, and 36
1. Find the LCM of 12, 15, and 36.
2. Divide the LCM by 100, which will give us a number that ends with two zeroes.
3. Subtract the remainder obtained when the above number is divided by 100 from the above number to get a number that is divisible by 100.
4. Divide the above number by 36 to get a number that is divisible by 36.
5. Check if the number obtained in step 4 is also divisible by 12 and 15.
6. If the number is divisible by 12 and 15, then it is the greatest number of 5 digits exactly divisible by 12, 15, and 36. Otherwise, subtract 36 from the number obtained in step 4 and repeat step 5 until we get a number that is divisible by 12 and 15.
Example
1. LCM of 12, 15, and 36 = 180
2. 180/100 = 1.8
3. 180 - 80 = 100
4. 100/36 = 2.7778
5. 2.7778 is not divisible by 12 and 15.
6. 2.7778 - 36 = -33.2222
7. -33.2222 + 36 = 2.7778
8. 2.7778 is divisible by 12 and 15.
9. Therefore, the greatest number of 5 digits exactly divisible by 12, 15, and 36 is 99,720.
Conclusion
In conclusion, to find the greatest number of 5 digits exactly divisible by 12, 15, and 36, we need to use the concept of Euclid's Division Lemma. We need to find the LCM of the given numbers, divide it by 100, subtract the remainder, divide by 36, and check if it is divisible by 12 and 15. If not, we need to subtract 36 and repeat the process until we get a number that is divisible by 12 and 15.
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