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61,121,181,241,301,361,421,481,541,601,661,721,781,841,901,961,1021,1081,1141,1201, ect

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Q: What number can be divided by 2 with a remainder of 1 divided by 3 with a remainder of 1 divided by 4 with a remainder of 1 divided by 5 with a remainder of 1?

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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.