3 divided by 2 has a remainder of 1. Which is 1 less than 2.

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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

solve it with a calculater

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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.

Put the remainder over the number you originally divided by. 15 divided by 7 = 2, remainder 1 = 2 and 1/7

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.

17

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.

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301 is one such number.

I think you're wanting a number of two digits, one of which is 3, that when divided by 7 gives a quotient and a remainder of 1 and when that quotient is divided by 2 it gives a remainder of 1: Answer: 36 36 ÷ 7 = 5 r 1 5 ÷ 2 = 2 r 1 If you want the number to be such that if it is divided by 7 the remainder is 1 and if it is divided by 2 the remainder is 1, then: Answer: 43 43 ÷ 7 = 6 r 1 43 ÷ 2 = 21 r 1

5

6

An even number can be divided by 2 evenly. An odd number will have a remainder of 1 when divided by 2.

85. A number when divided by 238, leaves a remainder 79. What will be the remainder when the number is divided bv 17 ? (1) 8 (2) 9 (3) 10 @) 11

7, 37, 67, 97, 127, and an infinite number of other numbers in this series (starting with 7 and increasing in increments of 30) have a remainder of 1 when divided by 6 and a remainder of 2 when divided by 5.

179 works until divided by 7 the remainder is 6. 2519 works till 10....

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Every 2 digit number, when divided by the number one less than it, will result in a remainder of 1.

An even number can be divided by 2 evenly. An odd number will have a remainder of 1 when divided by 2.

37

To convert any number in any base to another base, simply iteratively divide by the second base, using the rules of arithmetic for the first base, recording the remainders in reverse order, until the quotient is zero. For example, answering the question of how to convert 9310 to 10111012... 93 divided by 2 is 46 remainder 1 46 divided by 2 is 23 remainder 0 23 divided by 2 is 11 remainder 1 11 divided by 2 is 5 remainder 1 5 divided by 2 is 2 remainder 1 2 divided by 2 is 1 remainder 0 1 divided by 2 is 0 remainder 1 The answer, reading the remainders from bottom to top, is 10111012. This was not a good example, because the answer is palindromic, and can be read the same way forwards and backwards. Here is another example, converting 3710 into 1001012. 37 divided by 2 is 18 remainder 1 18 divided by 2 is 9 remainder 0 9 divided by 2 is 4 remainder 1 4 divided by 2 is 2 remainder 0 2 divided by 2 is 1 remainder 0 1 divided by 2 is 0 remainder 1 The answer is 1001012.