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There is no number that divides by 1 and leaves any remainder but 0.

But, if we exclude 1 from the list, 301 might be one answer. 721 is another. There are many more.

The least common multiple of 2, 3, 4, 5, 6 is 60. Therefore, (any multiple of 60) + 1 will work: 61, 121, 181, 241, etc.

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9y ago
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10y ago

Answer: 1081

we are solving the system of equations, well a system of linear congruences actually.

x=1 mod2

x=1 mod3

x=1 mod4

x=1 mod5

So we can use the Chinese Remainder Theorem. I won't include all the work but you get

Solutions: 1, 61 (mod 120)

Now you want a 4 digit number, so we use the fact that the LCM of 2,3,4 and 5 is 60

and add 60 however many times we need to get a 4 digit number

let's add 60x17=1020

the add 61 and we have 1081

Now let's check

if we divide any odd number by 2 remainder is 1

if we divide our number by 3 we have 360 so remander is 1

if we divide by 4 we have 270 with remainder 1

if we divide by 5 we have remainder 1 since those number divisible by 5 end in 0 or 5 and 1081 is 1 more than that.

so our answer works!

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or

Just use the Least Common Denominator for 2, 3, 4, and 5 and select the lowest multiple that has 4 digits. The LCD is 3 x 4 x 5 =60, which means that any number 1 more than a multiple of 60 will give you the above results. The first multiples of 60 with 4 digits are 1020 and 1080.

Both 1021 and 1081 will give you the remainder 1 for each division. So will any number that is 1 more than a 4 digit multiple of 60 (such as 6001).

1021 / 2 = 510 r. 1

1021 / 3 = 340 r. 1

1021 / 4 = 255 r. 1

1021 / 5 = 204 r. 1

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15y ago

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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14y ago

301: 7*43. 300 is divisible by 2, 3, 4, 5, and 6.

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15y ago

119

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14y ago

27

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Anonymous

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3y ago

12

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Anonymous

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3y ago
uh...................no iits not

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8

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Q: What 4 digit 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 and divided by 5 with a remainder of 1?
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