Question

find the greatest 6 digit number divisible by 28,49,63,and 70

Answer 1

Solution :-

prime factors of 28, 49, 63 and 70 :-

→ 28 = 2² * 7

→ 49 = 7²

→ 63 = 3² * 7

→ 70 = 2 * 5 * 7

LCM = 2² * 7² * 3² * 5 = 8820

now, we know that,

• greatest 6 digit number = 999999

so,

→ Dividing greatest 6 digit number by LCM we get,

• 999999 ÷ 8820
• Quotient = 113
• Remainder = 3339 .

therefore,

→ Required number = greatest 6 digit number - remainder = 999999 - 3339 = 996660 (Ans.)

Learn more :-

वह छोटी से छोटी संख्या बताईये जिसमे 7,9,11 से भाग देने पर 1,2,3 शेष बचे

https://brainly.in/question/9090122

Answer 2

Step-by-step explanation:

other way is to first take 7-digit smallest number and then dividing by L.C.M of the number

Then the questiont should be multiplied by the LCM