-3
-10
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.
A number is a multiple of 4 if the last 2 digits are a multiple of 4 The 10s digit is even and the last digit is 0, 4 or 8 The 10s digit is odd and the last digit is 2 or 6 A number is a multiple of 8 if the last 3 digits are a multiple of 8 The 100s digit is even and the last 2 digits are a multiple of 8 The 100s digit is odd and the last 2 digits are 4 times an odd number
14 is one of them
93
11 x 9 = 99
-3
738
-10
2
There are (99 - 9) = 90 2-digit numbers.13 of them are multiples of 7.The probability is 13/90 = 14.4% (rounded)
The largest two-digit multiple of six is 96.
97 is the largest 2-digit prime number.
No, in order for the number to be even, the last number must be a multiple of 2. It doesn't matter what the largest number is as long as the last number is a multiple of two.
97=largest 2 digit prime
8 x 12 which equals 96