mathschallenge.net logo

Remainder Of One

Problem

Find the smallest number, greater than 1, which has a remainder of 1 when divided by any of the numbers 2, 3, 4, 5, 6, or 7.


Solution

We note that 2times3 = 6 is the smallest number that divides evenly by 2 or 3. In the same way it should be clear that 2times3times5 = 30 will be the smallest number that divides by 2, 3, 5, or 6. Although it is obvious that divisibility by 7 requires a factor of 7, to ensure that it divides by 4 = 2times2, we only need add one extra factor of 2. Hence 2times3times5times2times7 = 420 is the smallest number that evenly divides by 2, 3, 4, 5, 6, or 7, and so 421 is the smallest number with a remainder of 1.

What is the smallest number which, when divided by any of the number from 2 to 10, has a remainder of 1?
What about the smallest number which when divided by 2 has a remainder of 1, when divided by 3 has a remainder of 2, or when divided by 4 has a remainder of 3?

Problem ID: 233 (31 Jul 2005)     Difficulty: 2 Star

Only Show Problem