Question
Download Solution PDFFind the remainder when the smallest 6-digit number divisible by 12, 15 and 25 is divided by 9.
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFGiven values: A number divisible by 12, 15, and 25
Concept:
The number is the least common multiple (LCM) of 12, 15, and 25. The remainder when a number is divided by 9 is the same as the remainder when the sum of its digits is divided by 9.
Calculation:
⇒ The LCM of 12, 15, and 25 is 300. The smallest 6-digit number divisible by 300 is 100200.
⇒ Sum of digits of 100200 = 1 + 0 + 0 + 2 + 0 + 0 = 3
⇒ Remainder when 3 is divided by 9 = 3
Therefore, the remainder when the smallest 6-digit number divisible by 12, 15, and 25 is divided by 9 is 3.
Alternate Method LCM of 12, 15 and 25 is = 300.
and
300 × 334 = 100200
So, the smallest 6-digit number which is divisible by 12, 15 and 25 is = 100200
When 100200 is divided by 9 then the remainder is:
100200 = 9 × 11133 + 3
remainder is = 3.
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