The largest 3 digit number which when divided by 69 and 12 to give a remainder of 1 is 829. What we want is one more than a common multiple of 69 and 12. The common multiples of 69 and 12 are the multiples of their least common multiple: 69 = 3 x 23 12 = 2^2 x 3 → lcm = 2^2 x 3 x 23 = 276 The largest multiple of 276 which is a 3 digit number: 999 ÷ 276 = 3 r 171 → largest 3 digit multiple of 276 is 3 × 276 = 828 → the required number is 828 + 1 = 829.
3 X 4 = 12
The greatest 2-digit even number is 98
-3
The answer is 66.
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The only positive two-digit multiple of 12 and 10 is 60.
12 and 72 24 and 36
2
The largest 3 digit number which when divided by 69 and 12 to give a remainder of 1 is 829. What we want is one more than a common multiple of 69 and 12. The common multiples of 69 and 12 are the multiples of their least common multiple: 69 = 3 x 23 12 = 2^2 x 3 → lcm = 2^2 x 3 x 23 = 276 The largest multiple of 276 which is a 3 digit number: 999 ÷ 276 = 3 r 171 → largest 3 digit multiple of 276 is 3 × 276 = 828 → the required number is 828 + 1 = 829.
3 X 4 = 12
-10
Since the greatest digit is 9 and the greatest 2-digit number is 99, the product of them is 891.
2 and 12, among others.
The greatest 2-digit even number is 98
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1 is the greatest common multiple of the numbers listed. 23 is a prime number & is only divisible by itself & 1. what is the greatest common multiple of 2?