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.
Yes- A remainder can be any number less than the dividend (the number by which the divisor is divided). An example of a 2 digit number is: 131/11=11 remainder 10.
300
You can't have a remainder of 6 when you divide by 2! JHC!
0.2308
27.2222
A 2 digit number divided by a four digit number, such as 2345, will leave the whole 2-digit number as a remainder. It cannot leave a remainder of 1.
Every 2 digit number, when divided by the number one less than it, will result in a remainder of 1.
The simplest answer is 10124 divided by 125.
17
11
103
12.5
100
111
Yes- A remainder can be any number less than the dividend (the number by which the divisor is divided). An example of a 2 digit number is: 131/11=11 remainder 10.
300
This is impossible. A two digit number n divided by 345 has a remainder equal to n. If you meant to say divided by 3,4,5 or 6 then the answer is 62.