The lowest number is 421.
There are many numbers that meet that criteria, all of them are one number higher than multiples of the Least Common Denominator of 2, 3, 4, 5, 6, and 7 (which is 3 x 4 x 5 x 7 = 420 ).
421 /2 = 210 r.1
421 /3 = 140 r.1
421 /4 = 105 r.1
421 /5 = 84 r.1
421 /6 = 70 r.1
421 /7 = 60 r.1
It is an integer which, when divided by 2, leaves a remainder of 1.
3 divided by 2 has a remainder of 1. Which is 1 less than 2.
29
59
solve it with a calculater
It is an integer which, when divided by 2, leaves a remainder of 1.
3 divided by 2 has a remainder of 1. Which is 1 less than 2.
29
59
solve it with a calculater
57
17
Put the remainder over the number you originally divided by. 15 divided by 7 = 2, remainder 1 = 2 and 1/7
No number can satisfy these conditions: To have a remainder of 1 when divided by 6, the number must be odd (as all multiples of 6 are even and an even number plus 1 is odd) To have a remainder of 2 when divided by 8, the number must be even (as all multiples of 8 are even and an even number plus 2 is even) No number is both odd and even. → No number exists that has a remainder of 1 when divided by 6, and 2 when divided by 8.
58
It is not possible, because the number 4 is divisible by 2, and it's remainder is divisible by 2 also, so whatever number works for the "4 with a remainder of 2", will never work for "2 with a remainder of 1.
The LCM of 2, 3, 4 and 5 is 60. Since you need a remainder of 1 just add 1. So the answer is 61. Or any number that is 1 more than a multiple of 60.